GATE EC 2017 Set 2 — Question 32

To find the Thevenin equivalent resistance $R_{th}$ across terminals P-Q, we deactivate all independent sources and apply a test source across P-Q. The 10V independent voltage source is shorted, which means the 1$\Omega$ resistor connected to it is now connected to ground.Let's label the nodes as follows: P (top), Q (bottom), A (left), and B (right). The circuit for finding $R_{th}$ is:

• A 1$\Omega$ resistor connects node A to ground.
• Resistors $R_{AP}$, $R_{AQ}$, $R_{PB}$, $R_{QB}$ are all 1$\Omega$.
• A dependent current source $3i_o$ is connected between A and B, with current flowing from B to A.
• The controlling current $i_o$ is the current through $R_{PB}$, flowing from P to B, so $i_o = (V_P - V_B)/1 = V_P - V_B$.

We apply a test voltage $V_t$ between P and Q, such that $V_P - V_Q = V_t$, and find the resulting current $I_t$ that flows from P to Q.Write KCL equations at nodes A, B, P, and Q.
KCL at A: $\frac{V_A - 0}{1} + \frac{V_A - V_P}{1} + \frac{V_A - V_Q}{1} - 3i_o = 0 \implies 3V_A - V_P - V_Q - 3(V_P - V_B) = 0 \implies 3V_A - 4V_P - V_Q + 3V_B = 0$ (1)
KCL at B: $\frac{V_B - V_P}{1} + \frac{V_B - V_Q}{1} + 3i_o = 0 \implies 2V_B - V_P - V_Q + 3(V_P - V_B) = 0 \implies -V_B + 2V_P - V_Q = 0 \implies V_B = 2V_P - V_Q$ (2)KCL at P (current leaving P is $I_t$): $I_t = \frac{V_P - V_A}{1} + \frac{V_P - V_B}{1} = 2V_P - V_A - V_B$ (3)
KCL at Q (current entering Q is $I_t$): $I_t = \frac{V_A - V_Q}{1} + \frac{V_B - V_Q}{1} = V_A + V_B - 2V_Q$ (4)From (3) and (4), we equate the expressions for $I_t$:
$2V_P - V_A - V_B = V_A + V_B - 2V_Q \implies 2V_P + 2V_Q = 2V_A + 2V_B \implies V_P + V_Q = V_A + V_B$ (5)Now substitute (2) into (1):
$3V_A - 4V_P - V_Q + 3(2V_P - V_Q) = 0 \implies 3V_A + 2V_P - 4V_Q = 0 \implies V_A = \frac{4V_Q - 2V_P}{3}$ (6)Now substitute (2) and (6) into (5):
$V_P + V_Q = \left(\frac{4V_Q - 2V_P}{3}\right) + (2V_P - V_Q)$
$V_P + V_Q = \frac{4V_Q - 2V_P + 6V_P - 3V_Q}{3} = \frac{4V_P + V_Q}{3}$
$3V_P + 3V_Q = 4V_P + V_Q \implies 2V_Q = V_P$.This is a constraint imposed by the circuit itself. Any excitation must satisfy $V_P = 2V_Q$. Our test source must respect this. Let $V_t = V_P - V_Q$. Substituting $V_P = 2V_Q$, we get $V_t = 2V_Q - V_Q = V_Q$. Thus, $V_Q = V_t$ and $V_P = 2V_t$.Now we can find the test current $I_t$ using equation (3):
$I_t = 2V_P - V_A - V_B$.
We need $V_A$ and $V_B$ in terms of $V_t$.
From (2): $V_B = 2V_P - V_Q = 2(2V_t) - V_t = 3V_t$.
From (6): $V_A = \frac{4V_Q - 2V_P}{3} = \frac{4(V_t) - 2(2V_t)}{3} = \frac{4V_t - 4V_t}{3} = 0$.Substitute these into the expression for $I_t$:
$I_t = 2(2V_t) - 0 - 3V_t = 4V_t - 3V_t = V_t$.Finally, the Thevenin resistance is:
$R_{th} = \frac{V_t}{I_t} = \frac{V_t}{V_t} = 1 \Omega$.The Thevenin equivalent resistance is 1 $\Omega$.