GATE EE 2023 Set 1 — Question 51

The circuit needs to be analyzed to find the Thevenin equivalent voltage $V_{ab}$ (open-circuit voltage) across terminals a-b.Given input current $i(t) = \sin(1000t)$.
The angular frequency is $\omega = 1000 \text{ rad/s}$.
In phasor form, $I = 1 \angle -90^{\circ} = -j$.Impedances:
Resistors: $R_1 = 10\Omega$ (left), $R_2 = 10\Omega$ (right)
Inductor: $L = 10\Omega$ (given as $j10\Omega$ in the diagram, so $Z_L = j10\Omega$)
Capacitor: $C = -j10\Omega$ (given as $-j10\Omega$ in the diagram, so $Z_C = -j10\Omega$)Let's assume the bottom wire is ground ($V_b = 0$). We need to find $V_a$. So, $i_x = (0 - V_2)/10 = -V_2/10$. So, $V_2 = I \cdot Z_L = i \cdot (j10) = j10i$. Therefore, the current flowing through the right $10\Omega$ resistor from $V_a$ to $V_3$ must be equal to the current $4i_x$.
$(V_a - V_3)/10 = 4i_x$
$V_a - V_3 = 40i_x$
$V_3 = V_a - 40i_x$. Current from $V_2$ to $V_3$ through $j10\Omega$ inductor: $I_{L} = (V_2 - V_3)/(j10)$
Current from $V_a$ to $V_3$ through $10\Omega$ resistor: $I_{R2} = (V_a - V_3)/10$
Current from $V_3$ to ground through $-j10\Omega$ capacitor: $I_{C} = V_3/(-j10)$
$I_{L} + I_{R2} + I_{C} = 0$
$\frac{V_2 - V_3}{j10} + \frac{V_a - V_3}{10} + \frac{V_3}{-j10} = 0$
Multiply by $j10$ to clear denominators:
$(V_2 - V_3) + j(V_a - V_3) - V_3 = 0$
$V_2 - V_3 + jV_a - jV_3 - V_3 = 0$
$V_2 + jV_a - (2+j)V_3 = 0$ $V_2 + jV_a - (2+j)(V_a - 40i_x) = 0$
$V_2 + jV_a - (2V_a - 80i_x + jV_a - j40i_x) = 0$
$V_2 + jV_a - 2V_a + 80i_x - jV_a + j40i_x = 0$
$V_2 - 2V_a + (80 + j40)i_x = 0$ $j10i - 2V_a + (80 + j40)(-ji) = 0$
$j10i - 2V_a - j80i - j^2 40i = 0$
$j10i - 2V_a - j80i + 40i = 0$
$-2V_a + (40 - j70)i = 0$
$2V_a = (40 - j70)i$
$V_a = (20 - j35)i$ $V_a = (20 - j35)(-j) = -j20 + j^2 35 = -j20 - 35$
So, the phasor voltage $V_a = -35 - j20 \text{ V}$. $V_a(t) = \text{Re}((-35 - j20)e^{j1000t})$
$V_a(t) = \text{Re}((-35 - j20)(\cos(1000t) + j\sin(1000t)))$
$V_a(t) = -35\cos(1000t) + 20\sin(1000t)$ $1000t = 1000 \times 5 \times 10^{-3} = 5 \text{ radians}$.
$V_a(5 \text{ ms}) = -35\cos(5) + 20\sin(5)$
Using a calculator (in radians):
$\cos(5 \text{ rad}) \approx 0.28366$
$\sin(5 \text{ rad}) \approx -0.95892$ $V_a(5 \text{ ms}) = -35(0.28366) + 20(-0.95892)$
$V_a(5 \text{ ms}) = -9.9281 - 19.1784$
$V_a(5 \text{ ms}) = -29.1065$Rounding off to 2 decimal places, the instantaneous value of the Thevenin's equivalent voltage is $-29.11 \text{ V}$.