GATE 2020 EE Q.32: solution and answer

GATE 2020 · EEQ.32 · 2 marks · MCQ

A benchtop dc power supply acts as an ideal 4 A current source as long as its terminal voltage is below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage source for all load currents going down to 0 A. When connected to an ideal rheostat, find the load resistance value at which maximum power is transferred, and the corresponding load voltage and current.

(A)
Short, $\infty$ A, 10 V
(B)
Open, 4 A, 0 V
(C)
2.5 $Ω$ , 4 A, 10 V
(D)
2.5 $Ω$ , 4 A, 5 V

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Answer and solution

ANSWER

The rheostat should be set to 2.5 Ω, where the load current is 4 A and the load voltage is 10 V, giving a maximum of 40 W. The answer is (C).

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

Supply characteristic. For $0 \leq V < 10 V$ the supply is an ideal current source, $I = 4 A$ . For $0 \leq I \leq 4 A$ it is an ideal voltage source, $V = 10 V$ . The two pieces meet at the corner $(V, I) = (10 V, 4 A)$ .
Corner load (Ohm's law). The load line $V = I R$ passes through that corner when $R = \frac{10 V}{4 A} = 2.5 Ω$
Current-source region, $R \leq 2.5 Ω$ . Here $V = 4 R \leq 10 V$ , so the supply holds $I = 4 A$ and $P = I^{2} R = 16 R$ Power rises with $R$ and reaches $16 \times 2.5 = 40 W$ at $R = 2.5 Ω$ .
Voltage-source region, $R \geq 2.5 Ω$ . Here $I = \frac{10}{R} \leq 4 A$ , so the supply holds $V = 10 V$ and $P = \frac{V ^{2}}{R} = \frac{100}{R}$ Power falls as $R$ increases, starting from $\frac{100}{2.5} = 40 W$ at $R = 2.5 Ω$ .
Maximum. $P$ rises up to $R = 2.5 Ω$ and falls beyond it, and both expressions give $40 W$ at the corner. So maximum power is delivered at $R = 2.5 Ω$ , where $I = 4 A$ , $V = 10 V$ and $P_{ma x} = 40 W$ .
Checking the other options. A short makes $V = 0$ and an open makes $I = 0$ , so both deliver zero power. Option (A) is also impossible as printed, because the supply limits its current to $4 A$ and a short cannot hold $10 V$ . Option (D) pairs $2.5 Ω$ and $4 A$ with $5 V$ , which breaks Ohm's law, since $4 \times 2.5 = 10 V$ . The answer is (C).

KEY IDEA

This supply has no linear Thevenin equivalent. It is current-limited below 10 V and voltage-limited up to 4 A, so the rule $R_{L} = R_{t h}$ does not apply. Delivered power $P = V I$ rises along the current-limited branch ( $P = 16 R$ ) and falls along the voltage-limited branch ( $P = 100/ R$ ), so the maximum is at the knee of the V–I characteristic, $(10 V, 4 A)$ , which the load reaches when $R = 10/4 = 2.5 Ω$ .

WATCH OUT FOR

Applying the linear maximum-power-transfer rule $R_{L} = R_{t h}$ . An ideal current source has $R_{t h} = \infty$ , which suggests an open load, and an ideal voltage source has $R_{t h} = 0$ , which suggests a short. Both loads deliver zero power here. A second trap is the matched-load habit of taking half the source voltage (5 V), which gives option (D). But 4 A through 2.5 Ω is 10 V, not 5 V, so (D) breaks Ohm's law.

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GATE 2020 EE Q.32: solution and answer

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