GATE 2022 EC Q.54: solution and answer

GATE 2022 · ECQ.54 · 2 marks · NAT

A linear 2-port network is shown in Fig. (a). An ideal DC voltage source of 10 V is connected across Port 1. A variable resistance $R$ is connected across Port 2. As $R$ is varied, the measured voltage and current at Port 2 is shown in Fig. (b) as a $V_{2}$ versus $- I_{2}$ plot. Note that for $V_{2} = 5 V$ , $I_{2} = 0 mA$ , and for $V_{2} = 4 V$ , $I_{2} = - 4 mA$ .

When the variable resistance $R$ at Port 2 is replaced by the load shown in Fig. (c), the current $I_{2}$ is ________ mA (rounded off to one decimal place).

GATE 2022 EC question 54: Fig. — Figure from the GATE 2022 EC paper. It is part of the question.

Answer and solution

ANSWER

$I_{2} = 4.0 mA$ .

Official answer: 3.9 to 4.1Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

Port 2 is a Thevenin source. The network is linear and port 1 is held at a fixed 10 V by an ideal source, so port 2 behaves as a linear one-port. Its $V_{2}$ – $I_{2}$ characteristic is a straight line, as Fig. (b) shows. At $I_{2} = 0$ (open circuit), $V_{t h} = V_{2} = 5 V$ , with + at A.
Thevenin resistance from the second point. $I_{2}$ is defined as flowing into the network at A, so $I_{2} = - 4 mA$ means 4 mA flows out of A into $R$ . That current drops $R_{t h} (- I_{2})$ across the Thevenin resistance, so $V_{2} = V_{t h} - R_{t h} (- I_{2})$ and $R_{t h} = \frac{5 - 4}{4 \times 1 0 ^{- 3}} = 250 Ω$ This is the magnitude of the slope of the line in Fig. (b).
Port-2 relation. With $I_{2}$ in amperes, taken into the network at A: $V_{2} = 5 + 250 I_{2}$ Check: $I_{2} = - 4 \times 1 0^{- 3}$ gives $V_{2} = 5 - 1 = 4 V$ , which matches Fig. (b).
Load relation from Fig. (c). The battery's positive (long) plate faces the resistor side and its negative plate is at B. The current that enters the network at A is the current that leaves the load at A, so it flows up through the load: through the battery from $-$ to $+$ (a 10 V rise), then through the $1 k Ω$ resistor towards A (a drop of $1000 I_{2}$ ). Hence $V_{2} = V_{A B} = 10 - 1000 I_{2}$
Solve the two equations (one KVL loop). $5 + 250 I_{2} = 10 - 1000 I_{2}$ $I_{2} = \frac{10 - 5}{1000 + 250} = \frac{5}{1250} = 4 \times 1 0^{- 3} A = 4.0 mA$
Check. $V_{2} = 5 + 250 (0.004) = 6 V$ and $10 - 1000 (0.004) = 6 V$ , so the two sides agree. $I_{2}$ comes out positive because the 10 V load battery is stronger than the 5 V Thevenin source, so it pushes current into the network at A. The answer is $I_{2} = 4.0 mA$ .

KEY IDEA

With port 1 held at a fixed voltage, a linear 2-port looks like a plain Thevenin source at port 2. The $V_{2}$ versus $- I_{2}$ line gives both parameters: its intercept is $V_{t h} = 5 V$ and the magnitude of its slope is $R_{t h} = 250 Ω$ . Once the Thevenin source is joined to the new load (1 kΩ in series with 10 V), the circuit is a single KVL loop: $I_{2} = (10 - 5) / (1000 + 250)$ .

WATCH OUT FOR

The usual mistake is getting the direction of $I_{2}$ wrong. $I_{2}$ is drawn into the network at A, so the given point $I_{2} = - 4 mA$ means current flows out of the network into $R$ . If you take the port relation as $V_{2} = 5 - 250 I_{2}$ , or add the load battery the wrong way round (treating the sources as series-aiding, for example $(10 + 5) /1250 = 12 mA$ ), you get the wrong sign or the wrong magnitude. A second mistake is to write $R_{t h} = 4 V /4 mA = 1 k Ω$ from the second point alone. That is the load resistance at that point, not the Thevenin resistance. $R_{t h}$ is the slope, $(5 - 4) /4 mA = 250 Ω$ .

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GATE 2022 EC Q.54: solution and answer

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