PYQs / GATE EE / 2017 / Set 2 / Q21 GATE EE 2017 Set 2 — Question 21 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +1 / -0 Medium Z, Y, h & ABCD Parameters Two-Port Networks Electric Circuits
Electric Circuits → Two-Port Networks → Z, Y, h & ABCD Parameters
Last updated 5 September 2026
Question For the given 2-port network, the value of transfer impedance
z 21 z_{21} z 21 in ohms is
______ .
Solution Transfer impedance
z 21 z_{21} z 21 is defined as
z 21 = V 2 I 1 ∣ I 2 = 0 z_{21} = \frac{V_2}{I_1} \Big|_{I_2=0} z 21 = I 1 V 2 I 2 = 0 .
Let the center node be
V c V_c V c . With port 2 open (
I 2 = 0 I_2 = 0 I 2 = 0 ), we apply nodal analysis.
At node 2:
V 2 − V 1 2 + V 2 − V c 2 = 0 ⟹ 2 V 2 − V 1 − V c = 0 ⟹ V c = 2 V 2 − V 1 \frac{V_2 - V_1}{2} + \frac{V_2 - V_c}{2} = 0 \implies 2V_2 - V_1 - V_c = 0 \implies V_c = 2V_2 - V_1 2 V 2 − V 1 + 2 V 2 − V c = 0 ⟹ 2 V 2 − V 1 − V c = 0 ⟹ V c = 2 V 2 − V 1 .
At node
V c V_c V c :
V c − V 1 4 + V c − V 2 2 + V c 2 = 0 \frac{V_c - V_1}{4} + \frac{V_c - V_2}{2} + \frac{V_c}{2} = 0 4 V c − V 1 + 2 V c − V 2 + 2 V c = 0 .
Multiplying by 4:
( V c − V 1 ) + 2 ( V c − V 2 ) + 2 V c = 0 ⟹ 5 V c − V 1 − 2 V 2 = 0 (V_c - V_1) + 2(V_c - V_2) + 2V_c = 0 \implies 5V_c - V_1 - 2V_2 = 0 ( V c − V 1 ) + 2 ( V c − V 2 ) + 2 V c = 0 ⟹ 5 V c − V 1 − 2 V 2 = 0 .
Substituting
V c = 2 V 2 − V 1 V_c = 2V_2 - V_1 V c = 2 V 2 − V 1 :
5 ( 2 V 2 − V 1 ) − V 1 − 2 V 2 = 0 ⟹ 10 V 2 − 5 V 1 − V 1 − 2 V 2 = 0 ⟹ 8 V 2 = 6 V 1 ⟹ V 1 = 4 3 V 2 5(2V_2 - V_1) - V_1 - 2V_2 = 0 \implies 10V_2 - 5V_1 - V_1 - 2V_2 = 0 \implies 8V_2 = 6V_1 \implies V_1 = \frac{4}{3}V_2 5 ( 2 V 2 − V 1 ) − V 1 − 2 V 2 = 0 ⟹ 10 V 2 − 5 V 1 − V 1 − 2 V 2 = 0 ⟹ 8 V 2 = 6 V 1 ⟹ V 1 = 3 4 V 2 .
The input current
I 1 I_1 I 1 is the sum of currents leaving node 1:
I 1 = V 1 − V 2 2 + V 1 − V c 4 = V 1 − V 2 2 + V 1 − ( 2 V 2 − V 1 ) 4 = V 1 − V 2 2 + 2 V 1 − 2 V 2 4 = V 1 − V 2 2 + V 1 − V 2 2 = V 1 − V 2 I_1 = \frac{V_1 - V_2}{2} + \frac{V_1 - V_c}{4} = \frac{V_1 - V_2}{2} + \frac{V_1 - (2V_2 - V_1)}{4} = \frac{V_1 - V_2}{2} + \frac{2V_1 - 2V_2}{4} = \frac{V_1 - V_2}{2} + \frac{V_1 - V_2}{2} = V_1 - V_2 I 1 = 2 V 1 − V 2 + 4 V 1 − V c = 2 V 1 − V 2 + 4 V 1 − ( 2 V 2 − V 1 ) = 2 V 1 − V 2 + 4 2 V 1 − 2 V 2 = 2 V 1 − V 2 + 2 V 1 − V 2 = V 1 − V 2 .
Substituting
V 1 = 4 3 V 2 V_1 = \frac{4}{3}V_2 V 1 = 3 4 V 2 :
I 1 = 4 3 V 2 − V 2 = 1 3 V 2 I_1 = \frac{4}{3}V_2 - V_2 = \frac{1}{3}V_2 I 1 = 3 4 V 2 − V 2 = 3 1 V 2 .
Therefore,
z 21 = V 2 I 1 = V 2 1 3 V 2 = 3 Ω z_{21} = \frac{V_2}{I_1} = \frac{V_2}{\frac{1}{3}V_2} = 3 \Omega z 21 = I 1 V 2 = 3 1 V 2 V 2 = 3Ω .
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Solution Transfer impedance
z 21 z_{21} z 21 is defined as
z 21 = V 2 I 1 ∣ I 2 = 0 z_{21} = \frac{V_2}{I_1} \Big|_{I_2=0} z 21 = I 1 V 2 I 2 = 0 .
Let the center node be
V c V_c V c . With port 2 open (
I 2 = 0 I_2 = 0 I 2 = 0 ), we apply nodal analysis.
At node 2:
V 2 − V 1 2 + V 2 − V c 2 = 0 ⟹ 2 V 2 − V 1 − V c = 0 ⟹ V c = 2 V 2 − V 1 \frac{V_2 - V_1}{2} + \frac{V_2 - V_c}{2} = 0 \implies 2V_2 - V_1 - V_c = 0 \implies V_c = 2V_2 - V_1 2 V 2 − V 1 + 2 V 2 − V c = 0 ⟹ 2 V 2 − V 1 − V c = 0 ⟹ V c = 2 V 2 − V 1 .
At node
V c V_c V c :
V c − V 1 4 + V c − V 2 2 + V c 2 = 0 \frac{V_c - V_1}{4} + \frac{V_c - V_2}{2} + \frac{V_c}{2} = 0 4 V c − V 1 + 2 V c − V 2 + 2 V c = 0 .
Multiplying by 4:
( V c − V 1 ) + 2 ( V c − V 2 ) + 2 V c = 0 ⟹ 5 V c − V 1 − 2 V 2 = 0 (V_c - V_1) + 2(V_c - V_2) + 2V_c = 0 \implies 5V_c - V_1 - 2V_2 = 0 ( V c − V 1 ) + 2 ( V c − V 2 ) + 2 V c = 0 ⟹ 5 V c − V 1 − 2 V 2 = 0 .
Substituting
V c = 2 V 2 − V 1 V_c = 2V_2 - V_1 V c = 2 V 2 − V 1 :
5 ( 2 V 2 − V 1 ) − V 1 − 2 V 2 = 0 ⟹ 10 V 2 − 5 V 1 − V 1 − 2 V 2 = 0 ⟹ 8 V 2 = 6 V 1 ⟹ V 1 = 4 3 V 2 5(2V_2 - V_1) - V_1 - 2V_2 = 0 \implies 10V_2 - 5V_1 - V_1 - 2V_2 = 0 \implies 8V_2 = 6V_1 \implies V_1 = \frac{4}{3}V_2 5 ( 2 V 2 − V 1 ) − V 1 − 2 V 2 = 0 ⟹ 10 V 2 − 5 V 1 − V 1 − 2 V 2 = 0 ⟹ 8 V 2 = 6 V 1 ⟹ V 1 = 3 4 V 2 .
The input current
I 1 I_1 I 1 is the sum of currents leaving node 1:
I 1 = V 1 − V 2 2 + V 1 − V c 4 = V 1 − V 2 2 + V 1 − ( 2 V 2 − V 1 ) 4 = V 1 − V 2 2 + 2 V 1 − 2 V 2 4 = V 1 − V 2 2 + V 1 − V 2 2 = V 1 − V 2 I_1 = \frac{V_1 - V_2}{2} + \frac{V_1 - V_c}{4} = \frac{V_1 - V_2}{2} + \frac{V_1 - (2V_2 - V_1)}{4} = \frac{V_1 - V_2}{2} + \frac{2V_1 - 2V_2}{4} = \frac{V_1 - V_2}{2} + \frac{V_1 - V_2}{2} = V_1 - V_2 I 1 = 2 V 1 − V 2 + 4 V 1 − V c = 2 V 1 − V 2 + 4 V 1 − ( 2 V 2 − V 1 ) = 2 V 1 − V 2 + 4 2 V 1 − 2 V 2 = 2 V 1 − V 2 + 2 V 1 − V 2 = V 1 − V 2 .
Substituting
V 1 = 4 3 V 2 V_1 = \frac{4}{3}V_2 V 1 = 3 4 V 2 :
I 1 = 4 3 V 2 − V 2 = 1 3 V 2 I_1 = \frac{4}{3}V_2 - V_2 = \frac{1}{3}V_2 I 1 = 3 4 V 2 − V 2 = 3 1 V 2 .
Therefore,
z 21 = V 2 I 1 = V 2 1 3 V 2 = 3 Ω z_{21} = \frac{V_2}{I_1} = \frac{V_2}{\frac{1}{3}V_2} = 3 \Omega z 21 = I 1 V 2 = 3 1 V 2 V 2 = 3Ω .
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