GATE 2023 EE Q.52: solution and answer

GATE 2023 · EEQ.52 · 2 marks · NAT

The admittance parameters of the passive resistive two-port network shown in the figure are
$y_{11} = 5 S, y_{22} = 1 S, y_{12} = y_{21} = - 2.5 S$
The power delivered to the load resistor $R_{L}$ in Watt is ________ (Round off to 2 decimal places).

GATE 2023 EE question 52: A box labelled 'Resistive Two-port Network' has input terminal 1 (top left), output terminal 2 (top right) and a common… — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The power delivered to the 6 Ω load is $P_{L} = \frac{115600}{486} \approx 237.86 W$ .

Official answer: 237 to 239Source: Official answer key, GATE 2023 EE (IIT Kanpur)

The 3 Ω resistor joins the top wire at port 1 to the top wire at port 2, so electrically it sits between terminals 1 and 2. On its own it is a two-port with the same port voltages as the given network, and its port currents add to the network's. That makes it a parallel connection. The given network and the resistor share the common terminal $1^{'}, 2^{'}$ , so this is a common-ground (three-terminal) parallel connection and adding the y-parameters is always valid. For a series element $G = \frac{1}{3} S$ between terminals 1 and 2: $y_{11} = y_{22} = \frac{1}{3} S$ and $y_{12} = y_{21} = - \frac{1}{3} S$ .
In a parallel connection the y-matrices add:
$Y = [5 + \frac{1}{3} - 2.5 - \frac{1}{3} - 2.5 - \frac{1}{3} 1 + \frac{1}{3}] = [\frac{16}{3} - \frac{17}{6} - \frac{17}{6} \frac{4}{3}] S$
Port conditions: the source fixes $V_{1} = 20 V$ . With $I_{2}$ taken as flowing into port 2, the load current flows out of port 2, so $I_{2} = - \frac{V _{2}}{R _{L}} = - \frac{V _{2}}{6}$ .
Substitute into the output equation $I_{2} = Y_{21} V_{1} + Y_{22} V_{2}$ :
$- \frac{V _{2}}{6} = - \frac{17}{6} (20) + \frac{4}{3} V_{2} \Rightarrow (\frac{4}{3} + \frac{1}{6}) V_{2} = \frac{340}{6} \Rightarrow \frac{3}{2} V_{2} = \frac{170}{3}$
Therefore $V_{2} = \frac{340}{9} \approx 37.78 V$ .
Check by nodal analysis at terminal 2. The network draws $y_{21} V_{1} + y_{22} V_{2} = - 50 + V_{2}$ , the 3 Ω draws $\frac{V _{2} - 20}{3}$ and the load draws $\frac{V _{2}}{6}$ . The three currents leaving the node sum to zero:
$(- 50 + V_{2}) + \frac{V _{2} - 20}{3} + \frac{V _{2}}{6} = 0 \Rightarrow 9 V_{2} = 340$
This gives the same $V_{2} = \frac{340}{9} V$ .
Power delivered to the load:
$P_{L} = \frac{V _{2}^{2}}{R _{L}} = \frac{( 340/9 ) ^{2}}{6} = \frac{115600}{486} \approx 237.86 W$

KEY IDEA

A resistor placed across the top terminals 1 and 2 of a common-ground two-port is itself a two-port connected in parallel with it. Add its y-matrix, $\frac{1}{R} [1 - 1 - 1 1]$ , to the given one. Then apply the two terminations: $V_{1}$ fixed by the source and $I_{2} = - V_{2} / R_{L}$ at the load.

WATCH OUT FOR

The most common mistake is to ignore the 3 Ω bridging resistor and use the given y-parameters alone. That gives $V_{2} = 300/7 V$ and about $306.12 W$ . Another mistake is to write the load condition as $I_{2} = + V_{2} / R_{L}$ , forgetting that $I_{2}$ is defined into the port. That gives $V_{2} = 340/7 V$ and about $393.20 W$ . Some students also give the bridging resistor $y_{12} = + \frac{1}{3} S$ instead of $- \frac{1}{3} S$ .

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2023 EE Q.52: solution and answer

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