GATE EC · EE · IN

Transients in RL, RC and RLC circuits

Transient analysis finds how the voltages and currents of a circuit with inductors and capacitors move from one steady state to another after a switch operates or a source is applied. A typical transient problem comes down to three steps. First, fix the initial conditions at $0^{+}$ from the continuity of capacitor voltage and inductor current. Second, find the final (or sinusoidal steady-state) value. Third, join the two through the circuit's natural response: a single time constant for a first-order circuit, or the roots of $s^{2} + 2 α s + ω_{0}^{2} = 0$ for a second-order one.

20 previous-year questions · 20 solved · 5 practice questions

Initial conditions: just before and just after switching

Conventions in these notes. A switch operates at $t = 0$ . The instant $t = 0^{-}$ is just before it operates (the old circuit, usually in DC steady state), and $t = 0^{+}$ is just after (the new circuit). Each element follows the passive sign convention: the reference current enters the terminal marked + for the reference voltage, so $v_{L} = L \frac{d i _{L}}{d t}$ and $i_{C} = C \frac{d v _{C}}{d t}$ . Sinusoidal amplitudes in these notes are peak values, not rms.

The energy stored in a capacitor, $\frac{1}{2} C v_{C}^{2}$ , and in an inductor, $\frac{1}{2} L i_{L}^{2}$ , cannot change in zero time unless infinite power is available. So as long as the capacitor current and the inductor voltage stay finite (no impulses), the capacitor voltage and the inductor current are continuous. Everything else can jump at the switching instant, including $i_{C}$ , $v_{L}$ and every resistor voltage and current.

CONTINUITY OF THE STATE VARIABLES

v_{C} (0^{+}) = v_{C} (0^{-}), i_{L} (0^{+}) = i_{L} (0^{-})

Holds whenever $i_{C}$ and $v_{L}$ contain no impulse. It can fail in two ways: a source that itself contains an impulse (such as $δ (t)$ ), or a switching that creates a loop of capacitors and ideal voltage sources whose voltages do not already satisfy KVL, or a cut-set of inductors and ideal current sources whose currents do not already satisfy KCL (see the tip below). Put in terms of charge and flux linkage: $q_{C} = C v_{C}$ and $λ_{L} = L i_{L}$ are continuous.

Element	At $t = 0^{+}$	In DC steady state ( $t \to \infty$ , or $t = 0^{-}$ after a long time)
Resistor	Resistor (its $v$ and $i$ may jump)	Resistor
Capacitor, $v_{C} (0^{-}) = 0$	Short circuit	Open circuit
Capacitor, $v_{C} (0^{-}) = V_{0}$	Voltage source $V_{0}$ (same polarity)	Open circuit
Inductor, $i_{L} (0^{-}) = 0$	Open circuit	Short circuit
Inductor, $i_{L} (0^{-}) = I_{0}$	Current source $I_{0}$ (same direction)	Short circuit

FIRST DERIVATIVES AT 0+

\frac{d v _{C}}{d t}_{0^{+}} = \frac{i _{C} ( 0 ^{+} )}{C}, \frac{d i _{L}}{d t}_{0^{+}} = \frac{v _{L} ( 0 ^{+} )}{L}

Find $i_{C} (0^{+})$ and $v_{L} (0^{+})$ from the $t = 0^{+}$ circuit, using KCL and KVL with capacitors replaced by voltage sources and inductors by current sources. A second-order circuit needs this as its second initial condition. For the derivative of a resistor quantity, differentiate its KVL or KCL equation and substitute these values.

Procedure for initial and final conditions

Draw the circuit for $t < 0$ . If it has been connected to DC sources for a long time, replace every capacitor by an open circuit and every inductor by a short circuit, then find every $v_{C} (0^{-})$ and $i_{L} (0^{-})$ .
Carry them over: $v_{C} (0^{+}) = v_{C} (0^{-})$ and $i_{L} (0^{+}) = i_{L} (0^{-})$ .
Draw the circuit for $t > 0$ as it is at $t = 0^{+}$ : each capacitor becomes a voltage source $v_{C} (0^{+})$ and each inductor a current source $i_{L} (0^{+})$ . The result is a resistive circuit. Solve it for any other voltage or current at $0^{+}$ .
From the same circuit, read $i_{C} (0^{+})$ and $v_{L} (0^{+})$ and divide by $C$ and $L$ to get the first derivatives.
For the final value, redraw the $t > 0$ circuit in DC steady state (capacitor open, inductor short).

EXAM TIP

Only $v_{C}$ and $i_{L}$ carry over from $0^{-}$ to $0^{+}$ . A common slip is to assume a resistor current, a capacitor current or an inductor voltage keeps its $0^{-}$ value. Recompute every one of them from the $0^{+}$ circuit.

EXAM TIP

Forced jumps. Suppose closing a switch puts a capacitor directly across an ideal voltage source, or directly across another capacitor at a different voltage. KVL then makes the two voltages equal at once, which takes an impulse of current, so $v_{C}$ jumps. Across an ideal voltage source, $v_{C} (0^{+})$ is simply the source voltage. Between capacitors, use conservation of charge: at a node that no voltage source touches, the total charge on the capacitor plates joined there is the same at $0^{+}$ as at $0^{-}$ (resistors pass no charge in zero time). The dual case is opening a switch that forces an inductor into series with an ideal current source, or two inductors carrying different currents into series. With a current source, $i_{L} (0^{+})$ is simply the source current. Between inductors, use conservation of flux linkage around a loop through them: with each current measured in the same direction around the loop, $\sum L_{k} i_{k}$ is the same at $0^{+}$ as at $0^{-}$ . Charge and flux linkage are conserved, but energy is not (see the energy section).

First-order RC and RL circuits

A circuit is first-order if, once its sources are set to zero, it reduces to one capacitor (or one inductor) plus a resistive network. Capacitors in series or parallel that combine into one count as one. The resistive part looks like a Thevenin (or Norton) equivalent from the element's terminals, so the capacitor or inductor always sees a single resistance $R_{t h}$ .

With DC sources, every voltage and current in the circuit then moves exponentially from its $0^{+}$ value to its final value, all with the same time constant.

TIME CONSTANT

τ = R_{t h} C (RC), τ = \frac{L}{R _{t h}} (RL)

$R_{t h}$ is the resistance seen from the capacitor or inductor terminals with independent voltage sources shorted and independent current sources opened. Keep dependent sources and find $R_{t h}$ with a test source, $R_{t h} = v_{t es t} / i_{t es t}$ . Use the circuit that exists after the switching.

COMPLETE RESPONSE OF A FIRST-ORDER CIRCUIT WITH DC SOURCES

x (t) = x (\infty) + [x (t_{0}^{+}) - x (\infty)] e^{- (t - t_{0}) / τ}, t \geq t_{0}

Valid for any voltage or current in a first-order circuit with constant sources. $t_{0}$ is the switching instant (usually $0$ ). Use the $0^{+}$ value, not the $0^{-}$ value, for anything other than $v_{C}$ and $i_{L}$ .

SERIES RC ON A DC SOURCE

v_{C} (t) = V_{s} + (V_{0} - V_{s}) e^{- t / R C}, i (t) = \frac{V _{s} - V _{0}}{R} e^{- t / R C}

DC source $V_{s}$ switched on at $t = 0$ with $v_{C} (0^{-}) = V_{0}$ . $i$ flows from the source into the + terminal of the capacitor. $V_{0} = 0$ gives charging, $V_{s} (1 - e^{- t / R C})$ ; $V_{s} = 0$ gives discharge, $V_{0} e^{- t / R C}$ .

SERIES RL ON A DC SOURCE

i_{L} (t) = \frac{V _{s}}{R} + (I_{0} - \frac{V _{s}}{R}) e^{- R t / L}, v_{L} (t) = (V_{s} - R I_{0}) e^{- R t / L}

DC source $V_{s}$ switched on at $t = 0$ with $i_{L} (0^{-}) = I_{0}$ . $I_{0} = 0$ gives build-up, $\frac{V _{s}}{R} (1 - e^{- R t / L})$ ; $V_{s} = 0$ gives decay, $I_{0} e^{- R t / L}$ . $v_{L}$ jumps to $V_{s} - R I_{0}$ at $t = 0^{+}$ ; if the inductor was in DC steady state before switching, $v_{L} (0^{-}) = 0$ .

INITIAL SLOPE AND TIME TO REACH A VALUE

\frac{d x}{d t}_{t_{0}^{+}} = \frac{x ( \infty ) - x ( t _{0}^{+} )}{τ}, t - t_{0} = τ ln \frac{x ( t _{0}^{+} ) - x ( \infty )}{x ( t ) - x ( \infty )}

The tangent at the start reaches the final value after exactly one time constant. The second form gives the instant at which $x$ reaches a stated value between $x (t_{0}^{+})$ and $x (\infty)$ .

RISE TIME AND HALF-CHANGE TIME

t_{10 - 90} = τ ln 9 \approx 2.2 τ, t_{50} = τ ln 2 \approx 0.693 τ

For a first-order step response. The rise time is the time to go from 10 to 90 percent of the change; $t_{50}$ is the time for half the change.

Time elapsed	Change completed, $1 - e^{- t / τ}$	Change remaining, $e^{- t / τ}$
$τ$	63.2%	36.8%
$2 τ$	86.5%	13.5%
$3 τ$	95.0%	5.0%
$4 τ$	98.2%	1.8%
$5 τ$	99.3%	0.7%

Switching sequences (more than one switch operation)

Solve interval 1 with the formula, using its own $x (\infty)$ and $τ$ .
At the next switching instant $t_{1}$ , evaluate the capacitor voltage or inductor current at $t_{1}^{-}$ from interval 1. By continuity this is the initial value $x (t_{1}^{+})$ for interval 2.
Redraw the circuit for interval 2. Find its new $x (\infty)$ and new $τ$ (the Thevenin resistance usually changes), and write the response with $(t - t_{1})$ in the exponent.
Repeat for each later switching. Solve for $v_{C}$ or $i_{L}$ first; the other quantities, which may jump at $t_{1}$ , then follow from the circuit at each instant.

WORKED EXAMPLE · EXAMPLE 1: A TWO-STEP SWITCHING SEQUENCE

A 10 V DC source is connected through a 5 kΩ resistor $R_{1}$ and a switch $S_{1}$ to an uncharged 2 µF capacitor. $S_{1}$ closes at $t = 0$ . At $t = 10$ ms a second switch $S_{2}$ closes, connecting a 5 kΩ resistor $R_{2}$ directly across the capacitor; $S_{1}$ stays closed. Find $v_{C}$ at $t = 10$ ms and at $t = 20$ ms, and the capacitor current just before and just after $t = 10$ ms.

Solution

Interval 1 ( $0 < t < 10$ ms): $v_{C} (0^{+}) = v_{C} (0^{-}) = 0$ , $v_{C} (\infty) = 10$ V and $τ_{1} = R_{1} C = 5 k Ω \times 2 μ F = 10$ ms, so $v_{C} (t) = 10 (1 - e^{- t /10 ms})$ V.
At $t = 1 0^{-}$ ms: $v_{C} = 10 (1 - e^{- 1}) = 6.321$ V and $i_{C} = \frac{10 - 6.321}{5 k Ω} = 0.736$ mA (charging).
Interval 2 ( $t > 10$ ms): the capacitor sees the source through the divider $R_{1}$ , $R_{2}$ . So $V_{t h} = 10 \times \frac{5}{5 + 5} = 5$ V and $R_{t h} = R_{1} ∥ R_{2} = 2.5$ kΩ, giving $v_{C} (\infty) = 5$ V and $τ_{2} = 2.5 k Ω \times 2 μ F = 5$ ms.
By continuity $v_{C} (1 0^{+}) = 6.321$ V, so for $t \geq 10$ ms, $v_{C} (t) = 5 + 1.321 e^{- (t - 10 ms) /5 ms}$ V.
$v_{C} (20 ms) = 5 + 1.321 e^{- 2} = 5 + 1.321 \times 0.1353 = 5.179$ V.
Just after $S_{2}$ closes, $i_{C} (1 0^{+}) = \frac{V _{t h} - v _{C} ( 1 0 ^{+} )}{R _{t h}} = \frac{5 - 6.321}{2.5 k Ω} = - 0.528$ mA. The capacitor current jumps from $+ 0.736$ mA to $- 0.528$ mA as the capacitor starts discharging into $R_{2}$ , while $v_{C}$ stays continuous.

Answer: $v_{C} (10 ms) = 6.32$ V and $v_{C} (20 ms) \approx 5.18$ V; $i_{C} (1 0^{-}) = + 0.736$ mA and $i_{C} (1 0^{+}) = - 0.528$ mA (charging direction taken as positive).

EXAM TIP

Four first-order traps. (1) $τ$ belongs to the circuit, not to one variable: within an interval, every voltage and current has the same $τ$ . (2) Find $R_{t h}$ from the circuit as it is after switching; a resistance that was switched out does not count. (3) Two capacitors separated by a resistor do not combine into one, and such a circuit is in general second-order. (4) With a dependent source, $R_{t h}$ can come out negative. A negative $τ$ means a growing exponential (an unstable circuit), and $x (\infty)$ is then not a final value the circuit reaches.

Second-order circuits: series and parallel RLC, and LC

A circuit with two energy-storage elements that cannot be combined is described by a second-order differential equation. For a series RLC circuit driven by $v_{s}$ , KVL gives $L C \frac{d ^{2} v _{C}}{d t ^{2}} + R C \frac{d v _{C}}{d t} + v_{C} = v_{s}$ . For a parallel RLC circuit driven by a current source $i_{s}$ , KCL gives $L C \frac{d ^{2} i _{L}}{d t ^{2}} + \frac{L}{R} \frac{d i _{L}}{d t} + i_{L} = i_{s}$ .

The complete response is the forced response (for DC sources, the constant final value) plus the natural response, whose form is set by the roots of the characteristic equation. The natural response has two constants, and they come from two initial conditions: $x (0^{+})$ and $\frac{d x}{d t} (0^{+})$ .

SERIES RLC: CHARACTERISTIC EQUATION

s^{2} + \frac{R}{L} s + \frac{1}{L C} = 0, α = \frac{R}{2 L}, ω_{0} = \frac{1}{L C}

For the source-free series loop (a voltage source is replaced by a short). $α$ is the damping coefficient (neper frequency, in s $^{- 1}$ ) and $ω_{0}$ is the undamped natural frequency (rad/s).

PARALLEL RLC: CHARACTERISTIC EQUATION

s^{2} + \frac{1}{R C} s + \frac{1}{L C} = 0, α = \frac{1}{2 R C}, ω_{0} = \frac{1}{L C}

For the source-free parallel circuit (a current source is replaced by an open). It has the same $ω_{0}$ as the series case but a different $α$ : here a larger $R$ means less damping.

ROOTS, DAMPING RATIO AND DAMPED FREQUENCY

s_{1, 2} = - α \pm α^{2} - ω_{0}^{2}, ζ = \frac{α}{ω _{0}}, ω_{d} = ω_{0}^{2} - α^{2} = ω_{0} 1 - ζ^{2}

Both cases have the form $s^{2} + 2 α s + ω_{0}^{2} = 0$ . $ω_{d}$ is real only when $α < ω_{0}$ . The roots multiply to $s_{1} s_{2} = ω_{0}^{2}$ and add to $- 2 α$ .

Case	Condition	Roots	Natural response $x_{n} (t)$
Overdamped	$α > ω_{0}$ ( $ζ > 1$ )	Real, distinct, negative: $s_{1}$ , $s_{2}$	$A_{1} e^{s_{1} t} + A_{2} e^{s_{2} t}$
Critically damped	$α = ω_{0}$ ( $ζ = 1$ )	Real, repeated: $- α$ , $- α$	$(A_{1} + A_{2} t) e^{- α t}$
Underdamped	$0 < α < ω_{0}$ ( $0 < ζ < 1$ )	Complex: $- α \pm j ω_{d}$	$e^{- α t} (B_{1} cos ω_{d} t + B_{2} sin ω_{d} t)$
Undamped (LC)	$α = 0$ ( $ζ = 0$ )	Imaginary: $\pm j ω_{0}$	$B_{1} cos ω_{0} t + B_{2} sin ω_{0} t$

CRITICAL RESISTANCE

R_{c, series} = 2 \frac{L}{C}, R_{c, parallel} = \frac{1}{2} \frac{L}{C}

A series RLC is overdamped for $R > 2 L / C$ and underdamped for $R < 2 L / C$ . A parallel RLC is the other way round: overdamped for $R < \frac{1}{2} L / C$ , underdamped for $R > \frac{1}{2} L / C$ .

QUALITY FACTOR

Q_{series} = \frac{1}{R} \frac{L}{C}, Q_{parallel} = R \frac{C}{L}, ζ = \frac{1}{2 Q}

Both equal $ω_{0} / (2 α)$ . Critical damping is $Q = \frac{1}{2}$ ; $Q > \frac{1}{2}$ is underdamped.

SERIES RLC SWITCHED ONTO A DC SOURCE: THE CURRENT

overdamped: critically damped: underdamped: i (t) = \frac{V _{s} - V _{0}}{L ( s _{1} - s _{2} )} (e^{s_{1} t} - e^{s_{2} t}) i (t) = \frac{V _{s} - V _{0}}{L} t e^{- α t} i (t) = \frac{V _{s} - V _{0}}{ω _{d} L} e^{- α t} sin ω_{d} t

DC source $V_{s}$ applied at $t = 0$ , with $i (0^{-}) = 0$ and $v_{C} (0^{-}) = V_{0}$ ; $i$ flows from the source into the + terminal of $C$ . All three cases follow from $i (0^{+}) = 0$ and $\frac{d i}{d t} (0^{+}) = \frac{V _{s} - V _{0}}{L}$ . With $R = 0$ : $i (t) = (V_{s} - V_{0}) C / L sin ω_{0} t$ .

SERIES RLC SWITCHED ONTO A DC SOURCE: THE CAPACITOR VOLTAGE

overdamped: critically damped: underdamped: v_{C} (t) = V_{s} + (V_{s} - V_{0}) \frac{s _{2} e ^{s_{1} t} - s _{1} e ^{s_{2} t}}{s _{1} - s _{2}} v_{C} (t) = V_{s} - (V_{s} - V_{0}) (1 + α t) e^{- α t} v_{C} (t) = V_{s} - (V_{s} - V_{0}) e^{- α t} (cos ω_{d} t + \frac{α}{ω _{d}} sin ω_{d} t)

Same conditions as the current formulas above. Each form satisfies $v_{C} (0^{+}) = V_{0}$ and $\frac{d v _{C}}{d t} (0^{+}) = i (0^{+}) / C = 0$ , and $i = C \frac{d v _{C}}{d t}$ gives back the matching current. With $R = 0$ : $v_{C} (t) = V_{s} - (V_{s} - V_{0}) cos ω_{0} t$ .

UNDERDAMPED CAPACITOR VOLTAGE: FIRST PEAK AND OVERSHOOT

t_{p} = \frac{π}{ω _{d}}, v_{C} (t_{p}) = V_{s} + (V_{s} - V_{0}) e^{- α π / ω_{d}}, \frac{v _{C} ( t _{p} ) - V _{s}}{V _{s} - V _{0}} = e^{- α π / ω_{d}} = e^{- ζ π / 1 - ζ^{2}}

Same conditions, with $V_{s} > V_{0}$ , so the first extremum is the maximum (for $V_{s} < V_{0}$ the same instant gives the lowest point, below $V_{s}$ ). The extrema of $v_{C}$ fall at $t = k π / ω_{d}$ ( $k = 1, 2, \dots$ ), the instants where $i = 0$ . Successive maxima are one damped period $2 π / ω_{d}$ apart, and the overshoot above $V_{s}$ shrinks by the factor $e^{- 2 π α / ω_{d}}$ from one maximum to the next. The fractional overshoot is the peak-overshoot formula of the standard second-order system.

LC CIRCUIT (UNDAMPED OSCILLATION)

v_{C} (t) = V_{0} cos ω_{0} t, i (t) = V_{0} \frac{C}{L} sin ω_{0} t, I_{m a x} = V_{0} \frac{C}{L}

A capacitor charged to $V_{0}$ is connected across an uncharged inductor at $t = 0$ ; $i$ is the current leaving the + terminal of $C$ and entering $L$ . The energy swings between $\frac{1}{2} C V_{0}^{2}$ and $\frac{1}{2} L I_{m a x}^{2}$ every quarter period. An uncharged LC switched onto a DC source $V_{s}$ gives $v_{C} = V_{s} (1 - cos ω_{0} t)$ , which peaks at $2 V_{s}$ .

Procedure for a second-order transient

Find $v_{C} (0^{+})$ and $i_{L} (0^{+})$ from the $t < 0$ circuit.
Choose the variable $x$ you need, and find $x (0^{+})$ and $\frac{d x}{d t} (0^{+})$ from the $0^{+}$ circuit.
Find the final value $x (\infty)$ from the DC steady state of the $t > 0$ circuit.
Set the sources to zero, find $α$ and $ω_{0}$ (or the characteristic equation), and pick the matching form from the table.
Write $x (t) = x (\infty) + x_{n} (t)$ and fit the two constants to $x (0^{+})$ and $\frac{d x}{d t} (0^{+})$ .

EXAM TIP

The characteristic equation belongs to the source-free circuit and is the same for every voltage and current in it, so compute it once. Do not mix up the series $α = R /2 L$ and the parallel $α = 1/ (2 R C)$ . For a circuit that is neither pure series nor pure parallel, write the loop or node equations in the $s$ -domain with the sources set to zero; the natural frequencies are the roots of their determinant. For a single loop, a quick route is to set the impedance seen at a break in the loop to zero, $Z (s) = 0$ . For a single node pair, set the admittance across it to zero, $Y (s) = 0$ .

EXAM TIP

Underdamped quantities overshoot. A de-energised series LC switched onto $V_{s}$ drives its capacitor voltage to a peak of $2 V_{s}$ (it then oscillates between $0$ and $2 V_{s}$ ), and a lightly damped RLC comes close to that. In a series RLC starting from rest, the overdamped and critically damped step responses of $v_{C}$ rise to $V_{s}$ without overshoot.

Laplace-domain analysis, impulse and step responses

The unilateral Laplace transform (lower limit $0^{-}$ ) turns each element's differential law into an algebraic one, and the initial conditions appear as extra sources. You can then solve the circuit with ordinary impedance methods (KVL, KCL, nodal, mesh, Thevenin, superposition) and invert the answer by partial fractions. Because it uses $0^{-}$ values, an impulse at $t = 0$ is handled automatically. Passive sign convention throughout.

RESISTOR IN THE S-DOMAIN

V_{R} (s) = R I_{R} (s)

Unchanged.

INDUCTOR WITH INITIAL CURRENT

V_{L} (s) = s L I_{L} (s) - L i_{L} (0^{-}) ⟺ I_{L} (s) = \frac{V _{L} ( s )}{s L} + \frac{i _{L} ( 0 ^{-} )}{s}

Series model: impedance $s L$ in series with a voltage source $L i_{L} (0^{-})$ that drives current in the direction of $i_{L}$ (its + terminal is on the side where $i_{L}$ leaves). Parallel model: impedance $s L$ in parallel with a current source $i_{L} (0^{-}) / s$ pointing in the direction of $i_{L}$ .

CAPACITOR WITH INITIAL VOLTAGE

V_{C} (s) = \frac{I _{C} ( s )}{s C} + \frac{v _{C} ( 0 ^{-} )}{s} ⟺ I_{C} (s) = s C V_{C} (s) - C v_{C} (0^{-})

Series model: impedance $\frac{1}{s C}$ in series with a voltage source $v_{C} (0^{-}) / s$ of the same polarity as $v_{C}$ . Parallel model: impedance $\frac{1}{s C}$ in parallel with a current source $C v_{C} (0^{-})$ whose arrow points toward the + terminal of $v_{C}$ (the charged capacitor pushes current out of its + terminal).

Procedure

Find $i_{L} (0^{-})$ and $v_{C} (0^{-})$ .
Transform the sources: a DC source $V$ switched on at $t = 0$ becomes $V / s$ ; $δ (t) \to 1$ ; $e^{- a t} u (t) \to \frac{1}{s + a}$ ; $sin (ω t) u (t) \to \frac{ω}{s ^{2} + ω ^{2}}$ ; $cos (ω t) u (t) \to \frac{s}{s ^{2} + ω ^{2}}$ .
Replace each L and C by its $s$ -domain model. The series (voltage-source) forms suit mesh analysis and the parallel (current-source) forms suit nodal analysis.
Solve for the unknown $X (s)$ , and check it with the initial- and final-value theorems.
Invert by partial fractions.

INITIAL- AND FINAL-VALUE THEOREMS

x (0^{+}) = s \to \infty lim s X (s), x (\infty) = s \to 0 lim s X (s)

For a rational $X (s)$ , the initial-value theorem needs $X (s)$ to be strictly proper; otherwise $x (t)$ has an impulse at $t = 0$ . The final-value theorem is valid only when every pole of $s X (s)$ lies in the open left half-plane. It gives a wrong answer for a sinusoid or a growing response.

NETWORK FUNCTION, IMPULSE RESPONSE AND STEP RESPONSE

H (s) = \frac{Y ( s )}{X ( s )}_{zero initial state}, h (t) = L^{- 1} {H (s)}, y_{step} (t) = L^{- 1} {\frac{H ( s )}{s}}, h (t) = \frac{d y _{step} ( t )}{d t}

Impulse and step responses are defined with every initial condition zero. The zero-state response to any input is $L^{- 1} {H (s) X (s)}$ , the convolution of $h (t)$ with the input. The complete response adds the zero-input response produced by the initial-condition sources. Every pole of $H (s)$ is a natural frequency of the circuit; a natural frequency can be missing from a particular $H (s)$ when it cancels against a zero, but it still appears in the zero-input response.

DELAYED INPUTS AND RECTANGULAR PULSES

f (t - T) u (t - T) \leftrightarrow e^{- s T} F (s), x (t) = V [u (t) - u (t - T)] \Rightarrow y (t) = V [y_{step} (t) - y_{step} (t - T)]

Time-shift property, with $T > 0$ . A pulse of height $V$ and width $T$ applied at zero initial state gives, by linearity and time invariance, the step response minus a copy delayed by $T$ (each copy is zero before its own step). For the series RC with output $v_{C}$ : $v_{C} = V (1 - e^{- t / R C})$ for $0 \leq t \leq T$ , and $v_{C} = V (1 - e^{- T / R C}) e^{- (t - T) / R C}$ for $t \geq T$ .

Circuit (input $v_{s}$ )	Output	$H (s)$	Unit-step response	Unit-impulse response
Series RC	$v_{C}$	$\frac{1}{1 + s R C}$	$(1 - e^{- t / R C}) u (t)$	$\frac{1}{R C} e^{- t / R C} u (t)$
Series RC	$i$	$\frac{s C}{1 + s R C}$	$\frac{1}{R} e^{- t / R C} u (t)$	$\frac{1}{R} δ (t) - \frac{1}{R ^{2} C} e^{- t / R C} u (t)$
Series RL	$i$	$\frac{1}{R + s L}$	$\frac{1}{R} (1 - e^{- R t / L}) u (t)$	$\frac{1}{L} e^{- R t / L} u (t)$
Series RL	$v_{L}$	$\frac{s L}{R + s L}$	$e^{- R t / L} u (t)$	$δ (t) - \frac{R}{L} e^{- R t / L} u (t)$
Series RLC	$v_{C}$	$\frac{ω _{0}^{2}}{s ^{2} + 2 α s + ω _{0}^{2}}$	From the second-order section with $V_{s} = 1$ , $V_{0} = 0$	$\frac{ω _{0}^{2}}{ω _{d}} e^{- α t} sin (ω_{d} t) u (t)$ (underdamped)

IMPULSES CHANGE THE STATE INSTANTLY

v_{C} (0^{+}) = v_{C} (0^{-}) + \frac{1}{C} \int_{0^{-}}^{0^{+}} i_{C} d t, i_{L} (0^{+}) = i_{L} (0^{-}) + \frac{1}{L} \int_{0^{-}}^{0^{+}} v_{L} d t

A current impulse $Q δ (t)$ into a capacitor raises its voltage by $Q / C$ . A voltage impulse $Λ δ (t)$ across an inductor raises its current by $Λ/ L$ . When the impulse is the only source, the circuit is source-free after $0^{+}$ , so the impulse response is the zero-input response from these new initial values.

WORKED EXAMPLE · EXAMPLE 2: SERIES RLC WITH AN INITIALLY CHARGED CAPACITOR

A series circuit with $R = 6 Ω$ , $L = 1$ H and $C = 0.125$ F is switched onto a 12 V DC source at $t = 0$ . The capacitor voltage $v_{C}$ is measured with the polarity the source charges it to. Just before switching, $i (0^{-}) = 0$ and $v_{C} (0^{-}) = 4$ V. Find $i (t)$ and $v_{C} (t)$ for $t > 0$ , and the maximum current.

Solution

Damping check: $α = \frac{R}{2 L} = 3 s^{- 1}$ and $ω_{0} = \frac{1}{L C} = 8 \approx 2.83$ rad/s. Since $α > ω_{0}$ the circuit is overdamped, with $s_{1, 2} = - 3 \pm 9 - 8 = - 2, - 4$ .
$s$ -domain loop: the source is $\frac{12}{s}$ , the resistor $6$ , the inductor $s L = s$ , and the capacitor $\frac{1}{s C} = \frac{8}{s}$ in series with its initial-condition source $\frac{4}{s}$ , which opposes the source around the loop. KVL: $\frac{12}{s} = I (s) (6 + s + \frac{8}{s}) + \frac{4}{s}$ .
$I (s) = \frac{8/ s}{6 + s + 8/ s} = \frac{8}{s ^{2} + 6 s + 8} = \frac{8}{( s + 2 ) ( s + 4 )} = \frac{4}{s + 2} - \frac{4}{s + 4}$ , so $i (t) = 4 (e^{- 2 t} - e^{- 4 t})$ A for $t > 0$ .
Cross-check with the second-order formula: $\frac{V _{s} - V _{0}}{L ( s _{1} - s _{2} )} = \frac{12 - 4}{1 \times ( - 2 + 4 )} = 4$ . It agrees.
$V_{C} (s) = \frac{I ( s )}{s C} + \frac{4}{s} = \frac{64}{s ( s + 2 ) ( s + 4 )} + \frac{4}{s} = \frac{8}{s} - \frac{16}{s + 2} + \frac{8}{s + 4} + \frac{4}{s}$ , so $v_{C} (t) = 12 - 16 e^{- 2 t} + 8 e^{- 4 t}$ V.
Checks: $v_{C} (0^{+}) = 12 - 16 + 8 = 4$ V (continuity), $v_{C} (\infty) = 12$ V, and $\frac{d v _{C}}{d t} (0^{+}) = 32 - 32 = 0 = i (0^{+}) / C$ .
Maximum current: $\frac{d i}{d t} = - 8 e^{- 2 t} + 16 e^{- 4 t} = 0 \Rightarrow e^{- 2 t} = \frac{1}{2} \Rightarrow t = \frac{l n 2}{2} \approx 0.347$ s, where $i_{m a x} = 4 (\frac{1}{2} - \frac{1}{4}) = 1$ A. At that instant $v_{C} = 12 - 8 + 2 = 6$ V.

Answer: $i (t) = 4 (e^{- 2 t} - e^{- 4 t})$ A and $v_{C} (t) = 12 - 16 e^{- 2 t} + 8 e^{- 4 t}$ V (overdamped); $i_{m a x} = 1$ A at $t = \frac{l n 2}{2} \approx 0.347$ s.

EXAM TIP

Two slips in $s$ -domain work. First, the inductor's series source $L i_{L} (0^{-})$ carries no $1/ s$ (it is the transform of the impulse $L i_{L} (0^{-}) δ (t)$ ), while the capacitor's series source is $v_{C} (0^{-}) / s$ . The parallel forms swap this: $i_{L} (0^{-}) / s$ for the inductor and $C v_{C} (0^{-})$ for the capacitor. Second, every initial-condition source must push current the way the stored energy would: the inductor's source keeps $i_{L}$ flowing in its old direction, and a charged capacitor's source drives current out of its + terminal.

AC (sinusoidal) excitation and the switching angle

When a sinusoidal source is switched on, the complete response is the sinusoidal steady state (found with phasors) plus a natural response. The natural response decays with the same time constant, or the same characteristic roots, as with a DC source. Its size depends on where in its cycle the source is at the switching instant.

Convention for this section. The source is $v (t) = V_{m} sin (ω t + θ)$ for $t > 0$ . $V_{m}$ is the peak value, and $θ$ (the switching angle) is the phase of the source at $t = 0$ . The circuit is de-energised at $0^{-}$ unless stated. If a source is written as $V_{m} cos (ω t + θ)$ , replace $θ$ by $θ + 9 0^{\circ}$ in the formulas below.

FIRST-ORDER CIRCUIT WITH ANY SOURCE

x (t) = x_{ss} (t) + [x (0^{+}) - x_{ss} (0^{+})] e^{- t / τ}

$x_{ss} (t)$ is the steady-state response (from phasors, for a sinusoid). This generalises the DC formula, where $x_{ss} = x (\infty)$ . The transient term vanishes when $x_{ss} (0^{+}) = x (0^{+})$ .

SERIES RL SWITCHED ONTO A SINUSOID

i (t) = \frac{V _{m}}{∣ Z ∣} [sin (ω t + θ - ϕ) - sin (θ - ϕ) e^{- R t / L}], ∣ Z ∣ = R^{2} + ω^{2} L^{2}, ϕ = tan^{- 1} \frac{ω L}{R}

With $i (0^{-}) = 0$ . The steady-state current lags the voltage by $ϕ$ . The exponential (DC offset) term starts at $- \frac{V _{m}}{∣ Z ∣} sin (θ - ϕ)$ .

SERIES RC SWITCHED ONTO A SINUSOID

v_{C} (t) = \frac{V _{m}}{ω C ∣ Z ∣} [- cos (ω t + θ + ϕ) + cos (θ + ϕ) e^{- t / R C}], ∣ Z ∣ = R^{2} + \frac{1}{ω ^{2} C ^{2}}, ϕ = tan^{- 1} \frac{1}{ω R C}

With $v_{C} (0^{-}) = 0$ . The steady-state current $\frac{V _{m}}{∣ Z ∣} sin (ω t + θ + ϕ)$ leads the voltage by $ϕ$ . The current's own transient term is $- \frac{V _{m}}{∣ Z ∣} \frac{c o s ( θ + ϕ )}{ω R C} e^{- t / R C}$ , which gives $i (0^{+}) = \frac{V _{m} s i n θ}{R}$ .

SWITCHING ANGLE FOR NO TRANSIENT

RL: θ = ϕ = tan^{- 1} \frac{ω L}{R}, RC: θ = 9 0^{\circ} - ϕ = tan^{- 1} (ω R C) (or 18 0^{\circ} more)

Both come from one rule: there is no transient when the steady-state value of the state variable ( $i_{L}$ or $v_{C}$ ) at the switching instant equals its actual initial value, which is zero here. For RL, switch when the steady-state current passes through zero; for RC, switch when the steady-state capacitor voltage does.

WORST-CASE SWITCHING FOR RL

θ - ϕ = \pm 9 0^{\circ} \Rightarrow offset = \mp \frac{V _{m}}{∣ Z ∣} e^{- R t / L}, i_{peak} \to \frac{2 V _{m}}{∣ Z ∣} when ω L ≫ R

The transient is largest when the switch closes at a peak of the steady-state current. About half a cycle later, the total current's first peak approaches twice the steady-state amplitude, provided the time constant is long compared with the period.

WORKED EXAMPLE · EXAMPLE 3: SWITCHING AN RL LOAD ONTO A SINUSOID

A de-energised series RL circuit has $R = 3 Ω$ and $X_{L} = ω L = 4 Ω$ at $ω = 100 π$ rad/s. The source $v (t) = 100 sin (100 π t + θ)$ V is switched on at $t = 0$ . (a) For which switching angles is there no transient? (b) Write $i (t)$ for $θ = 0$ and find $i$ at $t = 2$ ms. (c) What is the largest possible amplitude of the transient term, and at which $θ$ does it occur?

Solution

$∣ Z ∣ = 3^{2} + 4^{2} = 5 Ω$ , $ϕ = tan^{- 1} (4/3) = 53.1 3^{\circ}$ , and the steady-state amplitude is $V_{m} /∣ Z ∣ = 20$ A. $L = \frac{4}{100 π} = 12.73$ mH, so $τ = L / R = 4.244$ ms.
(a) There is no transient when $θ = ϕ = 53.1 3^{\circ}$ or $θ = 233.1 3^{\circ}$ : the steady-state current $20 sin (100 π t + θ - 53.1 3^{\circ})$ is then zero at $t = 0$ .
(b) With $θ = 0$ : $i (t) = 20 [sin (100 π t - 53.1 3^{\circ}) - sin (- 53.1 3^{\circ}) e^{- t / τ}] = 20 sin (100 π t - 53.1 3^{\circ}) + 16 e^{- t /4.244 ms}$ A. At $t = 0$ this gives $20 \times (- 0.8) + 16 = 0$ , as it must.
At $t = 2$ ms, $100 π t = 0.2 π$ rad $= 3 6^{\circ}$ . The sinusoid gives $20 sin (- 17.1 3^{\circ}) = - 5.890$ A and the exponential gives $16 e^{- 2/4.244} = 16 \times 0.6242 = 9.988$ A, so $i (2 ms) \approx 4.10$ A.
(c) The offset amplitude is $20 ∣ sin (θ - ϕ) ∣$ . It is largest, 20 A (the full steady-state amplitude), when $θ - ϕ = \pm 9 0^{\circ}$ , i.e. $θ = 143.1 3^{\circ}$ or $θ = - 36.8 7^{\circ}$ ( $323.1 3^{\circ}$ ).

Answer: (a) $θ = 53.1 3^{\circ}$ or $233.1 3^{\circ}$ . (b) $i (t) = 20 sin (100 π t - 53.1 3^{\circ}) + 16 e^{- t /4.244 ms}$ A, and $i (2 ms) \approx 4.10$ A. (c) 20 A, at $θ = 143.1 3^{\circ}$ or $323.1 3^{\circ}$ .

EXAM TIP

A de-energised series RLC circuit switched onto a sinusoid always has a transient. Avoiding one would need the steady-state current and the steady-state capacitor voltage both to be zero at $t = 0$ , but the two are $9 0^{\circ}$ apart and never zero together. For an RLC circuit with AC excitation, find the steady state with phasors, then fit the two constants of the natural response $x_{n}$ to $x_{n} (0^{+}) = x (0^{+}) - x_{ss} (0^{+})$ and $\frac{d x _{n}}{d t} (0^{+}) = \frac{d x}{d t} (0^{+}) - \frac{d x _{ss}}{d t} (0^{+})$ .

Energy during a transient

Energy bookkeeping gives quick answers and quick checks. Over a complete transient, the energy supplied by the sources equals the change in stored energy plus the energy dissipated in the resistors. Two results surprise students. The energy lost in charging a capacitor from a DC source does not depend on the resistance. And connecting two capacitors charged to different voltages loses energy, however small the connecting resistance.

STORED ENERGY

w_{C} = \frac{1}{2} C v_{C}^{2}, w_{L} = \frac{1}{2} L i_{L}^{2}

These depend only on the present $v_{C}$ or $i_{L}$ , which is why those two quantities are the continuous ones.

CHARGING A CAPACITOR FROM A DC SOURCE THROUGH A RESISTOR

W_{source} = C V_{s} (V_{s} - V_{0}), Δ W_{C} = \frac{1}{2} C (V_{s}^{2} - V_{0}^{2}), W_{R} = \frac{1}{2} C (V_{s} - V_{0})^{2}

From $v_{C} = V_{0}$ to the final value $V_{s}$ , through a resistor of any value. Starting uncharged, the source delivers $C V_{s}^{2}$ and exactly half of it is dissipated. If $V_{0} > V_{s}$ the capacitor discharges toward $V_{s}$ , and the same $W_{R}$ holds.

ENERGY DISSIPATED IN AN EXPONENTIAL DECAY

i (t) = I_{0} e^{- t / τ} \Rightarrow W_{R} (0 \to t) = \frac{1}{2} I_{0}^{2} R τ (1 - e^{- 2 t / τ})

The power $i^{2} R$ decays with time constant $τ /2$ . For an RL decay ( $τ = L / R$ ) the total is $\frac{1}{2} L I_{0}^{2}$ , and for an RC discharge from $V_{0}$ it is $\frac{1}{2} C V_{0}^{2}$ . After one time constant, $1 - e^{- 2} = 0.865$ of the stored energy has been dissipated.

CHARGE SHARING BETWEEN TWO CAPACITORS

V_{f} = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}, W_{lost} = \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}

For $C_{1}$ at $V_{1}$ and $C_{2}$ at $V_{2}$ connected in parallel, through a switch or a resistor. $V_{1}$ and $V_{2}$ are measured across the same pair of joined terminals, so a capacitor connected with reversed polarity enters with a negative voltage. The loss does not depend on the connecting resistance $R$ , which only sets the time constant $R \frac{C _{1} C _{2}}{C _{1} + C _{2}}$ .

FLUX SHARING BETWEEN TWO INDUCTORS

I_{f} = \frac{L _{1} I _{1} + L _{2} I _{2}}{L _{1} + L _{2}}, W_{lost} = \frac{1}{2} \frac{L _{1} L _{2}}{L _{1} + L _{2}} (I_{1} - I_{2})^{2}

For uncoupled inductors carrying $I_{1}$ and $I_{2}$ (measured in the same direction around the loop) that are forced into series at the switching instant. Flux linkage is conserved. The ideal model puts impulses of voltage across the inductors and the opening switch; in a real circuit the lost energy goes into the arc at the switch.

EXAM TIP

Energy checks. A source-free circuit whose natural response decays to zero dissipates exactly its initial stored energy by $t = \infty$ , whatever its damping. (The condition fails if some stored energy cannot reach a resistor, for example charge trapped on two capacitors in series, or current circulating in a loop of inductors with no resistance.) So the total heat in the resistor of a source-free series or parallel RLC circuit with $R > 0$ is $\frac{1}{2} C v_{C}^{2} (0^{+}) + \frac{1}{2} L i_{L}^{2} (0^{+})$ . If a question asks for the energy dissipated over a finite interval, integrate $i^{2} R$ over it; the stored-energy shortcut only works when the interval runs to $t = \infty$ .

PREVIOUS-YEAR QUESTIONS

20 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2018 · GATE EE 2026 · GATE EE 2025 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2020 · GATE EE 2018 · GATE IN 2025 · GATE IN 2024 · GATE IN 2023 · GATE IN 2022

GATE 2026 EC Q.23

GATE 2026 · ECQ.23 · 1 mark · MCQ

The relation between the input current ( $I$ ) and the output voltage ( $V$ ) of a circuit is governed by the equation: $C \frac{d V}{d t} = I (t) - m (t)$ . The circuit is excited by $I (t) = q δ (t)$ , where $q$ is a real valued constant. $V$ at $t = 0^{-}$ is $V_{0}$ .

Which of the following is an equivalent representation of the above case?

(A)
$C \frac{d V}{d t} = - m (t)$ , with $V (t = 0^{-}) = V_{0} + q / C$
(B)
$C \frac{d V}{d t} = - m (t)$ , with $V (t = 0^{-}) = V_{0} + q / C + m (t = 0^{-})$
(C)
$C \frac{d V}{d t} = - m (t)$ , with $V (t = 0^{-}) = V_{0} - q / C + m (t = 0^{-})$
(D)
$C \frac{d V}{d t} = - m (t)$ , with $V (t = 0^{-}) = q / C$

Answer and solution

GATE 2026 EC Q.23: answer and solution

See the question in the paper

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GATE 2026 EC Q.23: answer and solution

GATE 2026 EE Q.43

GATE 2026 · EEQ.43 · 2 marks · MCQ

The electrical network shown has an independent voltage source (10 V) and a current source (1 u(t) mA).

[Figure]

The voltage across the capacitor at time instants (in seconds) $t = 0^{+}$ , $t = 0.50$ , and $t = \infty$ , respectively, is:

GATE 2026 EE question 43: Single-node-pair circuit with a common bottom (reference) wire. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
8.00 V, 28.00 V, 26.36 V
(B)
8.00 V, 26.36 V, 28.00 V
(C)
10.00 V, 26.36 V, 28.00 V
(D)
10.00 V, 28.00 V, 26.36 V

Answer and solution

GATE 2026 EE Q.43: answer and solution

GATE 2025 EE Q.39

GATE 2025 · EEQ.39 · 2 marks · MCQ

In the circuit, shown below, if the values of R and C are very large, the form of the output voltage for a very high frequency square wave input, is best represented by

GATE 2025 EE question 39: Series RC low-pass network. — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
Figure: $V_{o u t}$ vs t is a triangular wave swinging the full range between $+ 1 V$ and $- 1 V$ . Its peaks and troughs touch the dashed 1 V and -1 V lines.
(B)
Figure: $V_{o u t}$ vs t is a square wave alternating between $+ 1 V$ and $- 1 V$ (the same as the input).
(C)
Figure: $V_{o u t}$ vs t is a small-amplitude triangular wave centred on the time axis (symmetric about 0 V). Its peaks sit only a fraction of a volt above and below the axis, far inside the dashed 1 V and -1 V lines.
(D)
Figure: $V_{o u t}$ vs t is a train of exponential spikes. It starts at $+ 1 V$ at t = 0 and decays to 0, stays at 0, jumps to $- 1 V$ at the next edge and decays back to 0, and so on alternately (a differentiated square wave).

Answer and solution

GATE 2025 EE Q.39: answer and solution

GATE 2025 EE Q.57

GATE 2025 · EEQ.57 · 2 marks · NAT

Q.57 The switch (S) closes at $t = 0$ sec. The time, in sec, the capacitor takes to charge to 50 V is _____ (round off to one decimal place).

GATE 2025 EE question 57: A single-loop circuit with a parallel group on its right. — Figure from the GATE 2025 EE paper. It is part of the question.

Answer and solution

GATE 2025 EE Q.57: answer and solution

GATE 2025 IN Q.63

GATE 2025 · INQ.63 · 2 marks · NAT

In the circuit shown, the switch is opened at $t = 0$ s. The current $i (t)$ at $t = 2$ ms is ________ mA (rounded off to two decimal places).

GATE 2025 IN question 63: A ladder-type circuit. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

GATE 2025 IN Q.63: answer and solution

GATE 2025 IN Q.64

GATE 2025 · INQ.64 · 2 marks · NAT

A signal $V_{M} = 5 sin (π t /3)$ V is applied to the circuit consisting of a switch $S$ and capacitor $C = 0.1 μ F$ , as shown in the figure. The output $V_{x}$ of the circuit is fed to an ADC having an input impedance consisting of a $10 M Ω$ resistance in parallel with a $0.1 μ F$ capacitor. If $S$ is opened at $t = 0.5$ s, the value of $V_{x}$ at $t = 1.5$ s will be ________ V (rounded off to two decimal places).

Note: Assume all components are ideal.

GATE 2025 IN question 64: Left: a plot of V_M (in V) against t (in s). — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

GATE 2025 IN Q.64: answer and solution

GATE 2024 EC Q.32

GATE 2024 · ECQ.32 · 1 mark · NAT

In the circuit given below, the switch S was kept open for a sufficiently long time and is closed at time $t = 0$ . The time constant (in seconds) of the circuit for $t > 0$ is ________.

GATE 2024 EC question 32: A ladder circuit on a common bottom (ground) rail. — Figure from the GATE 2024 EC paper. It is part of the question.

Answer and solution

GATE 2024 EC Q.32: answer and solution

GATE 2024 EE Q.14

GATE 2024 · EEQ.14 · 1 mark · MCQ

The circuit shown in the figure with the switch $S$ open, is in steady state. After the switch $S$ is closed, the time constant of the circuit in seconds is

GATE 2024 EE question 14: Two horizontal rails: a top wire and a bottom (reference) wire. — Figure from the GATE 2024 EE paper. It is part of the question.

(A)
1.25
(B)
0
(C)
1
(D)
1.5

Answer and solution

GATE 2024 EE Q.14: answer and solution

GATE 2024 IN Q.33

GATE 2024 · INQ.33 · 1 mark · NAT

The switch in the following figure has been closed for a long time ( $t < 0$ ). It is opened at $t = 0$ seconds. The value of $d v_{c} / d t$ at $t = 0^{+}$ is ________ V/s (rounded off to nearest integer).

GATE 2024 IN question 33: A single-loop-plus-branch circuit. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

GATE 2024 IN Q.33: answer and solution

GATE 2023 EC Q.35

GATE 2023 · ECQ.35 · 1 mark · NAT

In the circuit shown below, switch S was closed for a long time. If the switch is opened at $t = 0$ , the maximum magnitude of the voltage $V_{R}$ , in volts, is ________ (rounded off to the nearest integer).

GATE 2023 EC question 35: A single loop. — Figure from the GATE 2023 EC paper. It is part of the question.

Answer and solution

GATE 2023 EC Q.35: answer and solution

GATE 2023 EC Q.44

GATE 2023 · ECQ.44 · 2 marks · MCQ

The switch $S_{1}$ was closed and $S_{2}$ was open for a long time. At $t = 0$ , switch $S_{1}$ is opened and $S_{2}$ is closed, simultaneously. The value of $i_{c} (0^{+})$ , in amperes, is

GATE 2023 EC question 44: Parallel circuit with a common bottom rail and a top rail. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
1
(B)
$- 1$
(C)
0.2
(D)
0.8

Answer and solution

GATE 2023 EC Q.44: answer and solution

GATE 2023 EE Q.29

GATE 2023 · EEQ.29 · 1 mark · NAT

The value of parameters of the circuit shown in the figure are
$R_{1} = 2 Ω$ , $R_{2} = 2 Ω$ , $R_{3} = 3 Ω$ , $L = 10 mH$ , $C = 100 μ F$
For time $t < 0$ , the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at $t = 0$ , the value of the voltage across the inductor ( $V_{L}$ ) at $t = 0^{+}$ in Volts is ________ (Round off to 1 decimal place).

GATE 2023 EE question 29: A 10 A DC current source (arrow pointing up) is connected between a bottom wire (node B) and a top wire (node A). — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

GATE 2023 EE Q.29: answer and solution

GATE 2023 EE Q.54

GATE 2023 · EEQ.54 · 2 marks · NAT

The circuit shown in the figure is initially in the steady state with the switch $K$ in open condition and $\overline{K}$ in closed condition. The switch $K$ is closed and $\overline{K}$ is opened simultaneously at the instant $t = t_{1}$ , where $t_{1} > 0$ . The minimum value of $t_{1}$ in milliseconds, such that there is no transient in the voltage across the $100 μ F$ capacitor, is ________ (Round off to 2 decimal places).

GATE 2023 EE question 54: Three nodes: a top rail T, a bottom-left node M (drawn as a dot below switch K), and a bottom-right node R (a dot below… — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

GATE 2023 EE Q.54: answer and solution

GATE 2023 IN Q.47

GATE 2023 · INQ.47 · 2 marks · NAT

The R-L circuit with $R = 10 k Ω$ and $L = 1 mH$ is excited by a step current $I_{0} u (t)$ . At $t = 0^{-}$ , there is a current $I_{L} = I_{0} /5$ flowing through the inductor. The minimum time taken for the current through the inductor to reach 99% of its final value is ______ $μ s$ (rounded off to two decimal places).

GATE 2023 IN question 47: Three elements connected in parallel between a top rail and a bottom rail. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

GATE 2023 IN Q.47: answer and solution

GATE 2022 EE Q.31

GATE 2022 · EEQ.31 · 1 mark · NAT

In the circuit shown below, the switch $S$ is closed at $t = 0$ . The magnitude of the steady state voltage, in volts, across the $6 Ω$ resistor is ________. (round off to two decimal places).

GATE 2022 EE question 31: Three branches are connected in parallel between a left vertical rail and a right vertical rail. — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

GATE 2022 EE Q.31: answer and solution

GATE 2022 IN Q.62

GATE 2022 · INQ.62 · 2 marks · NAT

In the circuit shown, the capacitance $C_{0} = 10 μ F$ and inductance $L_{0} = 1 mH$ and the diode is ideal. The capacitor is initially charged to 10 V and the current in the inductor is initially zero. If the switch is closed at $t = 0$ s, the voltage $V_{c} (t)$ (in volts) across the capacitor at $t = 0.5$ s is ______ (round off to one decimal place)

GATE 2022 IN question 62: A single series loop. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

GATE 2022 IN Q.62: answer and solution

GATE 2022 IN Q.64

GATE 2022 · INQ.64 · 2 marks · NAT

In the circuit shown, the switch is initially closed. It is opened at $t = 0$ s and remains open thereafter. The time (in milliseconds) at which the output voltage $V_{o u t}$ becomes LOW is ________ (round off to three decimal places)

GATE 2022 IN question 64: A +5 V supply rail runs along the top. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

GATE 2022 IN Q.64: answer and solution

GATE 2020 EE Q.45

GATE 2020 · EEQ.45 · 2 marks · NAT

A resistor and a capacitor are connected in series to a 10 V dc supply through a switch. The switch is closed at $t = 0$ , and the capacitor voltage is found to cross 0 V at $t = 0.4 τ$ , where $τ$ is the circuit time constant. The absolute value of percentage change required in the initial capacitor voltage if the zero crossing has to happen at $t = 0.2 τ$ is ________ (rounded off to 2 decimal places).

Answer and solution

GATE 2020 EE Q.45: answer and solution

See the question in the paper

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GATE 2020 EE Q.45: answer and solution

GATE 2018 EC Q.36

GATE 2018 · ECQ.36 · 2 marks · NAT

For the circuit given in the figure, the magnitude of the loop current (in amperes, correct to three decimal places) 0.5 second after closing the switch is ________.

GATE 2018 EC question 36: A single series loop drawn as a rectangle. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

GATE 2018 EC Q.36: answer and solution

GATE 2018 EE Q.31

GATE 2018 · EEQ.31 · 2 marks · MCQ

A DC voltage source is connected to a series L-C circuit by turning on the switch S at time $t = 0$ as shown in the figure. Assume $i (0) = 0, v (0) = 0$ . Which one of the following circular loci represents the plot of $i (t)$ versus $v (t)$ ?

GATE 2018 EE question 31: Single series loop. — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
Figure — circle of radius 5 centred at $(0, - 5)$ in the $v$ – $i$ plane (below the $v (t)$ axis, tangent to it at the origin)
(B)
Figure — circle of radius 5 centred at $(5, 0)$ in the $v$ – $i$ plane (right of the $i (t)$ axis, tangent to it at the origin)
(C)
Figure — circle of radius 5 centred at $(0, 5)$ in the $v$ – $i$ plane (above the $v (t)$ axis, tangent to it at the origin)
(D)
Figure — circle of radius 5 centred at $(- 5, 0)$ in the $v$ – $i$ plane (left of the $i (t)$ axis, tangent to it at the origin)

Answer and solution

GATE 2018 EE Q.31: answer and solution

PRACTICE

5 questions to test yourself

PRACTICE 11 mark · NAT

A 10 V DC source, a switch S, a 2 kΩ resistor and a 5 µF capacitor are connected in series. Just before S closes at $t = 0$ , the capacitor voltage is 2 V, with the same polarity the source would charge it to. The capacitor voltage (in V, same polarity) at $t = 10$ ms is ______ (round off to two decimal places).

Answer and solution

Answer: $7.06$ (accepted range $\pm 0.02$ )

$v_{C} (0^{+}) = v_{C} (0^{-}) = 2$ V; $v_{C} (\infty) = 10$ V, since the capacitor is an open circuit in DC steady state.
$τ = R C = 2 k Ω \times 5 μ F = 10$ ms.
$v_{C} (10 ms) = 10 + (2 - 10) e^{- 1} = 10 - 8 \times 0.3679 = 7.057$ V $\approx 7.06$ V.

PRACTICE 21 mark · MCQ

A source-free parallel RLC circuit has $R = 2 Ω$ , $L = 4$ H and $C = \frac{1}{16}$ F. Its natural response is

(A)
underdamped, with $α = 0.25 s^{- 1}$ and $ω_{d} \approx 1.98$ rad/s
(B)
critically damped, with a repeated root at $s = - 2 s^{- 1}$
(C)
overdamped, with natural frequencies $\approx - 0.536 s^{- 1}$ and $\approx - 7.464 s^{- 1}$
(D)
undamped, oscillating at 2 rad/s

Answer and solution

Answer: (C)

$ω_{0} = \frac{1}{L C} = \frac{1}{4 \times 1/16} = 2$ rad/s.
For a parallel RLC, $α = \frac{1}{2 R C} = \frac{1}{2 \times 2 \times 1/16} = 4 s^{- 1}$ .
$α > ω_{0}$ , so the response is overdamped: $s_{1, 2} = - 4 \pm 16 - 4 = - 4 \pm 23 \approx - 0.536, - 7.464 s^{- 1}$ .
Option A is what the series formula $α = R /2 L = 0.25$ would give, and that formula does not apply to a parallel circuit.

PRACTICE 31 mark · MCQ

A 4 µF capacitor charged to 50 V and a 1 µF capacitor charged to 10 V are connected in parallel through a 100 Ω resistor at $t = 0$ , with their positive terminals joined. After the transient has died out, the common capacitor voltage and the total energy dissipated in the resistor are

(A)
42 V and 0.64 mJ
(B)
30 V and 0.64 mJ
(C)
42 V and 1.28 mJ
(D)
30 V and 1.28 mJ

Answer and solution

Answer: (A)

Charge is conserved: $V_{f} = \frac{4 \times 50 + 1 \times 10}{4 + 1} = \frac{210}{5} = 42$ V.
Energy lost $= \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2} = \frac{1}{2} \times 0.8 μ F \times 4 0^{2} = 0.64$ mJ.
Check: the initial energy is $\frac{1}{2} (4 μ F) (5 0^{2}) + \frac{1}{2} (1 μ F) (1 0^{2}) = 5.05$ mJ and the final energy is $\frac{1}{2} (5 μ F) (4 2^{2}) = 4.41$ mJ. The difference, 0.64 mJ, is all dissipated in the resistor, whatever its value.

PRACTICE 42 marks · MSQ

A 12 V DC source, a switch S and a 4 Ω resistor are connected in series between node a and the ground node g, with the source's + terminal toward a. Two more branches are connected between a and g: a 0.5 F capacitor, and a 2 H inductor in series with a 2 Ω resistor. S has been closed for a long time and opens at $t = 0$ , removing the source branch. Take $v_{C}$ as the voltage of node a with respect to g, and $i_{L}$ as the inductor current flowing from a toward g. Which of the following is/are correct?

(A)
$v_{C} (0^{+}) = 4$ V and $i_{L} (0^{+}) = 2$ A
(B)
$\frac{d v _{C}}{d t}$ at $t = 0^{+}$ is $- 4$ V/s
(C)
$\frac{d i _{L}}{d t}$ at $t = 0^{+}$ is $2$ A/s
(D)
For $t > 0$ the natural response is underdamped, with damped natural frequency $\frac{3}{2}$ rad/s

Answer and solution

Answer: (A), (B), (D)

For $t < 0$ , in DC steady state, the capacitor is open and the inductor is a short. The source drives $\frac{12}{4 + 2} = 2$ A through the 4 Ω resistor, the inductor and the 2 Ω resistor, so $i_{L} (0^{-}) = 2$ A. The capacitor is across the inductor branch, so $v_{C} (0^{-}) = 2 \times 2 = 4$ V.
By continuity, $v_{C} (0^{+}) = 4$ V and $i_{L} (0^{+}) = 2$ A. (A) is correct.
For $t > 0$ the capacitor and the inductor branch form a single loop. KCL at node a, with $i_{C}$ flowing from a to g through C: $i_{C} + i_{L} = 0$ , so $i_{C} (0^{+}) = - 2$ A and $\frac{d v _{C}}{d t} (0^{+}) = \frac{- 2}{0.5} = - 4$ V/s. (B) is correct.
$v_{L} (0^{+}) = v_{C} (0^{+}) - 2 i_{L} (0^{+}) = 4 - 4 = 0$ , so $\frac{d i _{L}}{d t} (0^{+}) = 0$ . (C) is wrong.
The loop is a series RLC with $R = 2 Ω$ , $L = 2$ H and $C = 0.5$ F: $α = \frac{R}{2 L} = 0.5 s^{- 1}$ and $ω_{0} = \frac{1}{L C} = 1$ rad/s. It is underdamped with $ω_{d} = 1 - 0.25 = \frac{3}{2} \approx 0.866$ rad/s. (D) is correct.

PRACTICE 52 marks · NAT

The voltage $v (t) = 200 sin (100 t)$ V ( $t$ in seconds, argument in radians) is switched on at $t = 0$ across a series circuit of $R = 10 Ω$ and $L = \frac{3}{10}$ H, which carries no current before switching. The current $i$ (in A) at $t = \frac{π}{200}$ s ( $\approx 15.71$ ms) is ______ (round off to two decimal places).

Answer and solution

Answer: $8.5$ (accepted range $\pm 0.03$ )

$ω L = 100 \times \frac{3}{10} = 103 Ω$ , so $∣ Z ∣ = 100 + 300 = 20 Ω$ , $ϕ = tan^{- 1} 3 = \frac{π}{3}$ rad $(6 0^{\circ})$ and $V_{m} /∣ Z ∣ = 10$ A.
$τ = \frac{L}{R} = \frac{3}{100}$ s $\approx 17.32$ ms.
With switching angle $θ = 0$ : $i (t) = 10 [sin (100 t - \frac{π}{3}) - sin (- \frac{π}{3}) e^{- t / τ}] = 10 [sin (100 t - \frac{π}{3}) + 0.8660 e^{- t / τ}]$ A.
At $t = π /200$ s, $100 t = \frac{π}{2}$ , so $sin (\frac{π}{2} - \frac{π}{3}) = sin \frac{π}{6} = 0.5$ . Also $\frac{t}{τ} = \frac{π /200}{3 /100} = \frac{π}{2 3} = 0.9069$ , $e^{- 0.9069} = 0.4038$ , and $0.8660 \times 0.4038 = 0.3497$ .
$i = 10 (0.5 + 0.3497) \approx 8.50$ A.

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