GATE 2023 EE Q.51: solution and answer

GATE 2023 · EEQ.51 · 2 marks · NAT

For the circuit shown, if $i = sin 1000 t$ , the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminals $a$ - $b$ at time $t = 5$ ms is ______ (Round off to 2 decimal places).

GATE 2023 EE question 51: Three parallel paths between a top rail and a bottom rail. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The Thevenin voltage is $v_{t h} (t) = 12.5 sin (1000 t) V$ , so at $t = 5 ms$ it is $12.5 sin (5 rad) = - 11.99 V$ .

Official answer: -12.1 to -11.9Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Work in phasors at $ω = 1000 rad/s$ , taking the source as the sine reference with its peak value: $I = 1∠ 0^{\circ} A$ (peak). The reactances are given directly on the figure. With $a$ - $b$ open, no current leaves through terminal $a$ , so $V_{t h} = V_{ab}$ is the voltage at node A (the top of the $10 - j 10 Ω$ branch).
Branch 1 (Ohm's law): $V_{A} = i_{x} (10 - j 10)$ .
Dependent source (+ at A, - at B): $V_{A} - V_{B} = 4 i_{x} \Rightarrow V_{B} = i_{x} (10 - j 10 - 4) = i_{x} (6 - j 10)$ .
KCL at node B: the source current splits between the $10 + j 10 Ω$ branch and the path through the dependent source to node A. Because terminal $a$ is open, all the current reaching node A must go down branch 1, so the dependent-source path carries exactly $i_{x}$ : $I = \frac{V _{B}}{10 + j 10} + i_{x} = i_{x} [\frac{6 - j 10}{10 + j 10} + 1]$
Simplify: $\frac{6 - j 10}{10 + j 10} = \frac{( 6 - j 10 ) ( 10 - j 10 )}{200} = \frac{60 - 100 - j 160}{200} = - 0.2 - j 0.8$ . So $I = i_{x} (0.8 - j 0.8)$ and $i_{x} = \frac{I}{0.8 ( 1 - j )} = \frac{( 1 + j ) I}{1.6} = 0.625 (1 + j) I$ .
Thevenin voltage: $V_{t h} = i_{x} (10 - j 10) = 0.625 (1 + j) \cdot 10 (1 - j) I = 6.25 \times 2 I = 12.5 I$ This is purely real, so $V_{t h} = 12.5∠ 0^{\circ} V$ (peak) is in phase with the source current. (Check: $V_{B} = i_{x} (6 - j 10) = 10 - j 2.5$ , and $\frac{V _{B}}{10 + j 10} + i_{x} = 1$ , as required.)
Back to the time domain: $v_{t h} (t) = 12.5 sin (1000 t) V$ .
At $t = 5 ms$ the argument is $1000 t = 5 rad$ (radians, not degrees): $v_{t h} = 12.5 sin (5) = 12.5 \times (- 0.95892) = - 11.99 V$ .

KEY IDEA

With $a$ - $b$ open, $V_{t h}$ is just the open-circuit voltage at node A. Because no current leaves through terminal $a$ , the current through the dependent source equals the controlling current $i_{x}$ . That turns the circuit into one KCL equation at node B with one unknown, $i_{x}$ . The resulting transfer $V_{t h} / I = 12.5$ is purely real, so the Thevenin voltage is in phase with $sin 1000 t$ .

WATCH OUT FOR

Working out $sin (5^{\circ})$ with the calculator in degree mode gives $+ 1.09 V$ . The argument $1000 t = 5$ is in radians, so the answer is $- 11.99 V$ . Other slips: writing the dependent-source KVL with the polarity reversed ( $V_{B} = V_{A} + 4 i_{x}$ ), and assuming the source path carries none of the current or all of it, when with $a$ - $b$ open it carries exactly $i_{x}$ .

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GATE 2023 EE Q.51: solution and answer

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