Laws & node/mesh formulas

Circuit laws, sources, and node and mesh analysis. 32 formulas for GATE EC, EE, IN Networks (Electric Circuits), each with when to use it. Each is explained in the topic notes.

All 32 Laws & node/mesh formulas

RESISTOR (OHM'S LAW)

v = R i, p = v i = i^{2} R = \frac{v ^{2}}{R} \geq 0

Linear resistor, passive sign convention. Conductance $G = 1/ R$ (siemens), so $i = G v$ .

INDUCTOR

v i (t) w_{L} = L \frac{d i}{d t} = i (t_{0}) + \frac{1}{L} \int_{t_{0}}^{t} v d τ = \frac{1}{2} L i^{2}

Passive sign convention. The stored energy depends only on the present current. In DC steady state $d i / d t = 0$ , so an inductor behaves as a short circuit.

CAPACITOR

i v (t) q = C \frac{d v}{d t} = v (t_{0}) + \frac{1}{C} \int_{t_{0}}^{t} i d τ = C v, w_{C} = \frac{1}{2} C v^{2} = \frac{q ^{2}}{2 C}

Passive sign convention. In DC steady state $d v / d t = 0$ , so a capacitor behaves as an open circuit.

CONTINUITY OF CAPACITOR VOLTAGE AND INDUCTOR CURRENT

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

$0^{-}$ is the instant just before the switching and $0^{+}$ the instant just after. The rule holds whenever the capacitor current and the inductor voltage stay finite (no impulses). That is guaranteed when the sources are finite and the switching creates no loop made only of capacitors and ideal voltage sources and no cut-set made only of inductors and ideal current sources. If such a loop is formed, the capacitor voltages are forced to jump (solve with charge conservation); if such a cut-set is formed, the inductor currents are forced to jump (solve with flux-linkage conservation). Capacitor current, inductor voltage and resistor quantities CAN jump at $t = 0$ .

SERIES AND PARALLEL COMBINATIONS

Series: R = R_{1} + R_{2}, L = L_{1} + L_{2}, \frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} Parallel: \frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}}, \frac{1}{L} = \frac{1}{L _{1}} + \frac{1}{L _{2}}, C = C_{1} + C_{2}

Inductors without mutual coupling. Two resistors in parallel give $R_{1} R_{2} / (R_{1} + R_{2})$ , and two capacitors in series give $C_{1} C_{2} / (C_{1} + C_{2})$ . Series capacitors that start uncharged carry equal charge.

COUPLED INDUCTORS (DOT CONVENTION)

v_{1} v_{2} M = L_{1} \frac{d i _{1}}{d t} \pm M \frac{d i _{2}}{d t} = L_{2} \frac{d i _{2}}{d t} \pm M \frac{d i _{1}}{d t} = k L_{1} L_{2}, 0 \leq k \leq 1

Each coil uses the passive sign convention for its own $v$ and $i$ . Take +M when both currents enter (or both leave) the dotted terminals, and −M when one enters a dot and the other leaves one. $k = 1$ means perfect coupling.

COUPLED INDUCTORS IN SERIES AND PARALLEL; STORED ENERGY

Series: L_{e q} = L_{1} + L_{2} \pm 2 M Parallel: L_{e q} = \frac{L _{1} L _{2} - M ^{2}}{L _{1} + L _{2} \mp 2 M} Energy: w = \frac{1}{2} L_{1} i_{1}^{2} + \frac{1}{2} L_{2} i_{2}^{2} \pm M i_{1} i_{2}

For $L_{e q}$ : upper sign = aiding (in series, the current enters both dotted terminals in turn; in parallel, the dotted terminals are joined together), lower sign = opposing. For the energy: $+ M i_{1} i_{2}$ when $i_{1}$ and $i_{2}$ both enter (or both leave) the dotted terminals, $- M i_{1} i_{2}$ otherwise. $w \geq 0$ for all currents, which is why $M \leq L_{1} L_{2}$ .

COUPLED COILS WITH A COMMON TERMINAL (T-EQUIVALENT)

Like terminals joined: L_{A} = L_{1} - M, L_{B} = L_{2} - M, L_{C} = M Unlike terminals joined: L_{A} = L_{1} + M, L_{B} = L_{2} + M, L_{C} = - M

Two coupled coils that share one terminal can be replaced by an uncoupled T network: series arm $L_{A}$ on coil 1's side, series arm $L_{B}$ on coil 2's side, and shunt arm $L_{C}$ to the common terminal. "Like terminals joined" means both dotted ends (or both undotted ends) meet at the common terminal. A negative inductance in the model is only a mathematical element; the behaviour at the three terminals is exact.

CHARGE SHARING BETWEEN TWO CAPACITORS

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

Capacitors charged to $V_{1}$ and $V_{2}$ are connected in parallel, + plate to + plate. If they are joined with opposite polarities, replace $V_{2}$ by $- V_{2}$ . The total charge is conserved; the energy $W_{lost}$ is dissipated in the connecting resistance, whatever its value.

SOURCE TRANSFORMATION

I_{s} = \frac{V _{s}}{R}, V_{s} = I_{s} R (same R)

$V_{s}$ in series with $R$ is equivalent at its two terminals to $I_{s}$ in parallel with the same $R$ . The current-source arrow points toward the terminal that was + on the voltage source. The two are equivalent only at the terminals: the power dissipated inside differs. An ideal source (no $R$ ) cannot be transformed. A dependent source can be transformed the same way, provided its controlling variable is not the current or voltage of the resistor $R$ being transformed, because that quantity changes in the transformation.

KIRCHHOFF'S CURRENT LAW

k \sum i_{k} = 0 (currents leaving a node or any closed surface)

Equivalently, total current entering = total current leaving. It also holds for a closed surface enclosing several nodes, which is the basis of the supernode.

KIRCHHOFF'S VOLTAGE LAW

k \sum v_{k} = 0 (voltage drops around any closed loop)

Count a drop as positive when the loop crosses an element from + to −. It holds for any closed path, including one through a current source or across an open gap (the voltage there is then an unknown).

VOLTAGE DIVISION

V_{k} = V \frac{R _{k}}{R _{1} + R _{2} + \dots + R _{N}}

Resistors in series carrying the same current, with nothing else drawing current from the tap. For impedances, replace $R$ by $Z$ .

CURRENT DIVISION

I_{1} = I \frac{R _{2}}{R _{1} + R _{2}}, I_{k} = I \frac{G _{k}}{G _{1} + G _{2} + \dots + G _{N}}

Branches in parallel across the same voltage. With two branches, the current in one is proportional to the OTHER branch's resistance.

BALANCED WHEATSTONE BRIDGE

\frac{R _{A C}}{R _{C B}} = \frac{R _{A D}}{R _{D B}} \Rightarrow V_{C} = V_{D}, I_{C D} = 0

A bridge with arms A–C–B and A–D–B and an element between C and D. When the bridge is balanced and the C–D element contains no source, that element carries no current and can be removed or replaced by a short without changing anything else.

CUBE OF TWELVE EQUAL RESISTORS

R_{body diagonal} = \frac{5}{6} R, R_{face diagonal} = \frac{3}{4} R, R_{edge} = \frac{7}{12} R

Resistance between two corners of a cube whose twelve edges are each $R$ : opposite corners of the cube, opposite corners of one face, and the two ends of one edge. Each follows from symmetry: join the corners that are at equal potential, then reduce.

INFINITE LADDER NETWORK

R_{in} = \frac{R _{1} + R _{1}^{2} + 4 R _{1} R _{2}}{2} (R_{1} = R_{2} = R : R_{in} = \frac{1 + 5}{2} R \approx 1.618 R)

A ladder that starts at the input with a series resistor $R_{1}$ , then a shunt resistor $R_{2}$ , repeated without end. Adding one more section does not change the input resistance, so $R_{in} = R_{1} + (R_{2} ∥ R_{in})$ , and the formula is the positive root. If the ladder starts with the shunt arm instead, its input resistance is $R_{2} ∥ R_{in}$ with $R_{in}$ from the formula.

DELTA TO STAR

R_{a} R_{b} R_{c} = \frac{R _{ab} R _{c a}}{R _{ab} + R _{b c} + R _{c a}} = \frac{R _{ab} R _{b c}}{R _{ab} + R _{b c} + R _{c a}} = \frac{R _{b c} R _{c a}}{R _{ab} + R _{b c} + R _{c a}}

Star arm at a node = product of the two delta resistors touching that node ÷ sum of all three delta resistors.

STAR TO DELTA

R_{ab} S = \frac{S}{R _{c}}, R_{b c} = \frac{S}{R _{a}}, R_{c a} = \frac{S}{R _{b}} = R_{a} R_{b} + R_{b} R_{c} + R_{c} R_{a}

Delta resistor between two nodes = sum of the pairwise products of the star arms ÷ the star arm at the OPPOSITE (third) node. Equivalently, $R_{ab} = R_{a} + R_{b} + R_{a} R_{b} / R_{c}$ .

BALANCED (EQUAL-ARM) CASE

R_{Y} = \frac{R _{Δ}}{3}, R_{Δ} = 3 R_{Y}

All three arms equal. The same relation holds for equal impedances at a single frequency, for example a balanced three-phase load.

NODAL EQUATIONS BY INSPECTION

[G] [V] = [I] G_{k k} = sum of conductances connected to node k G_{k j} = - (sum of conductances between nodes k and j)

Valid for circuits with only resistors and independent current sources. $I_{k}$ = net source current injected INTO node $k$ . The matrix is symmetric; a dependent source breaks the symmetry and has to be added by hand.

MESH EQUATIONS BY INSPECTION

[R] [I] = [V] R_{k k} = sum of resistances in mesh k R_{k j} = - (sum of resistances shared by meshes k and j)

Valid when all mesh currents are clockwise and the circuit has only resistors and independent voltage sources. $V_{k}$ = algebraic sum of the source voltages driving current in mesh $k$ 's direction (rises positive). The matrix is symmetric unless dependent sources are present.

NUMBER OF KCL OR KVL EQUATIONS TO WRITE

N_{nodal} N_{mesh} = (n - 1) - n_{V} = (b - n + 1) - n_{I}

$n$ = nodes and $b$ = branches of a connected circuit, counting every element as a separate branch. $n_{V}$ = voltage sources (independent or dependent); each fixes a node voltage or supplies a supernode constraint, so one KCL equation fewer is needed. $n_{I}$ = current sources (independent or dependent); each fixes a mesh current or supplies a supermesh constraint. A dependent source's constraint is still expressed in the unknowns, so it has to be substituted, not just read off. Assumes no loop made only of voltage sources and no cut-set made only of current sources. Choose the method that needs fewer equations.

POWER, ENERGY AND POWER BALANCE

p = v i, W = \int_{t_{1}}^{t_{2}} p d t, all elements \sum p_{absorbed} = 0

Passive sign convention. $1 kWh = 3.6 \times 1 0^{6}$ J.

SAFE CURRENT AND VOLTAGE OF A RATED RESISTOR

I_{m a x} = \frac{P _{rated}}{R}, V_{m a x} = P_{rated} R

The power rating is the largest power a component can dissipate continuously without overheating. In series, the smallest $I_{m a x}$ limits the whole string; in parallel, the smallest $V_{m a x}$ limits.

EFFICIENCY

η = \frac{P _{out}}{P _{in}} = \frac{P _{out}}{P _{out} + P _{loss}}

Losses are the power dissipated in resistances, for example $I^{2} R$ in conductors and in a source's internal resistance.

PRACTICAL SOURCE FEEDING A RESISTIVE LOAD

I η = \frac{V _{s}}{R _{s} + R _{L}}, P_{L} = \frac{V _{s}^{2} R _{L}}{( R _{s} + R _{L} ) ^{2}} = \frac{R _{L}}{R _{s} + R _{L}}

$η$ is the fraction of the power generated by the ideal part, $V_{s} I$ , that reaches $R_{L}$ . It approaches 100% as $R_{L} ≫ R_{s}$ .

MAXIMUM POWER FROM A PRACTICAL SOURCE

R_{L} = R_{s} \Rightarrow P_{L, m a x} = \frac{V _{s}^{2}}{4 R _{s}}, η = 50%

Fixed $V_{s}$ and $R_{s}$ with a variable resistive load. Maximum power transfer and maximum efficiency happen at different operating points. If instead $R_{L}$ is fixed and $R_{s}$ can be chosen, $P_{L} = V_{s}^{2} R_{L} / (R_{s} + R_{L})^{2}$ is largest at $R_{s} = 0$ , not at $R_{s} = R_{L}$ : the matching condition applies only when the load is the variable.

BATTERY WITH INTERNAL RESISTANCE

V_{t} V_{t} = E - I r (discharging) = E + I r (charging)

$E$ = EMF (the open-circuit voltage, which depends on the state of charge), $r$ = internal resistance, $I$ = magnitude of the current. Short-circuit current is $E / r$ . Stored energy ≈ nominal voltage × capacity in Ah, which gives Wh.

SOLAR CELL (IDEAL SINGLE-DIODE MODEL)

I I_{sc} V_{oc} = I_{p h} - I_{0} (e^{V / (n V_{T})} - 1) = I_{p h} = n V_{T} ln (1 + \frac{I _{p h}}{I _{0}})

Generator convention: $I$ is the current delivered out of the + terminal. $I_{p h}$ is the photocurrent (proportional to irradiance), $I_{0}$ the diode saturation current, $n$ the ideality factor, and $V_{T} = k T / q \approx 25.9$ mV at 300 K. Series and shunt resistances are neglected.

TWIGS AND LINKS OF A CONNECTED GRAPH

twigs = n - 1, links = b - n + 1

For a graph with $p$ separate parts: twigs $= n - p$ and links $= b - n + p$ .

INDEPENDENT EQUATIONS

N_{KCL} = n - 1, N_{KVL} = b - n + 1

Connected graph. For a planar graph, $b - n + 1$ is also the number of meshes (windows).

Where this was asked: 2018 (3), 2020 (4), 2022 (3), 2023 (2), 2024 (5), 2025 (3), 2026 (5). See the 25 previous-year questions on Laws & node/mesh. This shows what was asked, not what will be asked.

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Laws & node/mesh: the notes and 25 solved previous-year questions