GATE 2018 EE Q.19: solution and answer

GATE 2018 · EEQ.19 · 1 mark · NAT

In the two-port network shown, the $h_{11}$ parameter $(where, h_{11} = \frac{V _{1}}{I _{1}}, when V_{2} = 0)$ in ohms is ________ (up to 2 decimal places).

GATE 2018 EE question 19: A two-port network with a common bottom (ground) rail running from the port-1 minus terminal to the port-2 minus terminal. — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

ANSWER

$h_{11} = 0.50 Ω$

Official answer: 0.45 to 0.55Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Apply the test condition. $h_{11}$ is measured with $V_{2} = 0$ , so short port 2: node B (the port-2 + terminal) is now at the potential of the bottom rail. Let $V_{A}$ be the voltage of the central node A with respect to the bottom rail.
KCL at node A (current in = current out): $I_{1}$ enters through the input 1 Ω resistor. Current leaves through the 1 Ω shunt ( $V_{A} /1$ ), through the 1 Ω resistor to the shorted node B ( $(V_{A} - 0) /1$ ), and through the dependent source, which carries $2 I_{1}$ from A towards B: $I_{1} = \frac{V _{A}}{1} + \frac{V _{A} - 0}{1} + 2 I_{1}$
Solve for the node voltage: $2 V_{A} = I_{1} - 2 I_{1} = - I_{1} \Rightarrow V_{A} = - \frac{I _{1}}{2}$ The node voltage is negative because the dependent source takes more current out of node A ( $2 I_{1}$ ) than $I_{1}$ brings in, so the two resistors must return the difference to node A.
KVL around the input loop (port-1 + terminal, through the 1 Ω resistor, to node A, then to the bottom rail): $V_{1} = 1 \cdot I_{1} + V_{A} = I_{1} - \frac{I _{1}}{2} = 0.5 I_{1}$
The h-parameter: $h_{11} = \frac{V _{1}}{I _{1}}_{V_{2} = 0} = 0.50 Ω$

KEY IDEA

$h_{11}$ is the input impedance with the output shorted. Set $V_{2} = 0$ so that node B is grounded, write one KCL equation at the central node with the controlled source current $2 I_{1}$ included, and then $V_{1} = 1 \cdot I_{1} + V_{A}$ . You cannot simplify the network with series/parallel reduction because the controlled source depends on $I_{1}$ . The dependent source lowers the input impedance from the passive value $1 + (1 ∥ 1) = 1.5 Ω$ to $0.5 Ω$ .

WATCH OUT FOR

Ignoring the dependent source, or treating it as disappearing when port 2 is shorted, and reporting the passive value $1 + (1 ∥ 1) = 1.5 Ω$ . With $V_{2} = 0$ the source is not removed. It still takes $2 I_{1}$ out of node A (into the short), so it must stay in the KCL equation. A second slip is getting its direction wrong (taking $2 I_{1}$ as entering node A). That gives $V_{A} = 1.5 I_{1}$ and $h_{11} = 2.5 Ω$ .

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2018 EE Q.19: solution and answer

Answer and solution

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