GATE 2018 EC Q.25: solution and answer

GATE 2018 · ECQ.25 · 1 mark · NAT

The $A B C D$ matrix for a two-port network is defined by:

[V_{1} I_{1}] = [A C B D] [V_{2} - I_{2}]

The parameter $B$ for the given two-port network (in ohms, correct to two decimal places) is _______.

GATE 2018 EC question 25: A T-network two-port. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

ANSWER

$B = \frac{24}{5} = 4.80 Ω$ .

Official answer: 4.3 to 5.3Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Definition of $B$ . From $V_{1} = A V_{2} + B (- I_{2})$ , setting $V_{2} = 0$ gives $B = \frac{V _{1}}{- I _{2}}_{V_{2} = 0}$ . So short-circuit port 2 and find the ratio of $V_{1}$ to the current $- I_{2}$ that flows out through the short.
Input resistance with port 2 shorted. With port 2 shorted, the right-hand $2 Ω$ runs from the central node to the bottom line, so it sits in parallel with the $5 Ω$ shunt: $5 ∥ 2 = \frac{5 \times 2}{5 + 2} = \frac{10}{7} Ω$ . Hence $R_{in} = 2 + \frac{10}{7} = \frac{24}{7} Ω$ and $I_{1} = \frac{V _{1}}{R _{in}} = \frac{7 V _{1}}{24}$ .
Current divider at the central node. $I_{1}$ splits between the $5 Ω$ shunt and the right-hand $2 Ω$ . The share that goes through the $2 Ω$ and out of port 2 through the short is $- I_{2} = I_{1} \cdot \frac{5}{5 + 2} = \frac{7 V _{1}}{24} \cdot \frac{5}{7} = \frac{5 V _{1}}{24}$ .
Compute $B$ . $B = \frac{V _{1}}{- I _{2}} = \frac{V _{1}}{5 V _{1} /24} = \frac{24}{5} = 4.80 Ω$
Cross-check with the T-network formula. For series arms $Z_{1}$ , $Z_{2}$ and shunt arm $Z_{3}$ , $B = Z_{1} + Z_{2} + \frac{Z _{1} Z _{2}}{Z _{3}} = 2 + 2 + \frac{2 \times 2}{5} = 4.80 Ω$ . This agrees with the direct calculation.

KEY IDEA

$B$ is the transfer resistance $V_{1} / (- I_{2})$ with the output shorted ( $V_{2} = 0$ ). With port 2 shorted, find $I_{1}$ from the input resistance, then use a current divider to get the current through the short. For a T-network, $B = Z_{1} + Z_{2} + Z_{1} Z_{2} / Z_{3}$ gives the answer in one line.

WATCH OUT FOR

Taking $B$ to be the input resistance with port 2 shorted, $V_{1} / I_{1} = 24/7 \approx 3.43 Ω$ , which falls outside the key. $B$ divides $V_{1}$ by the output short-circuit current $- I_{2}$ , not by $I_{1}$ . Some candidates also open-circuit port 2 (that condition defines $A$ and $C$ , not $B$ ), or get the divider backwards and use $2/7$ in place of $5/7$ .

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

GATE 2018 EC Q.25: solution and answer

Answer and solution

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