GATE 2023 IN Q.33: solution and answer

GATE 2023 · INQ.33 · 1 mark · NAT

The diode in the circuit is ideal. The current source $i_{s} (t) = π sin (3000 π t)$ mA. The magnitude of the average current flowing through the resistor R is ______ mA (rounded off to two decimal places).

GATE 2023 IN question 33: Three elements connected in parallel between a top node and a bottom node: (1) on the left, an independent current source… — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

The magnitude of the average current through R is 1.00 mA.

Official answer: 0.95 to 1.05Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Circuit. The source, the ideal diode and R are all in parallel between a top node and a bottom node. The diode's anode is on the top node and its cathode is on the bottom node. The source arrow points up, so a positive $i_{s}$ pushes current into the top node. Take $i_{R}$ as the current flowing down through R, from the top node to the bottom node.
Positive half-cycle ( $i_{s} > 0$ ): the current entering the top node can flow through the diode from anode to cathode, which is its forward direction, so the diode is ON. An ideal ON diode is a short circuit, so the voltage across R is $0$ and $i_{R} = 0$ . All of $i_{s}$ flows through the diode. This is self-consistent because the diode current $i_{s}$ is positive in its forward direction.
Negative half-cycle ( $i_{s} < 0$ ): the source now pulls current out of the top node. If the diode were ON it would have to carry this current from cathode to anode, which an ideal diode cannot do, so the diode is OFF (open). The source and R then form a single loop, so $i_{R} = i_{s} = π sin (3000 π t) mA < 0$ . Check: the voltage of the top node relative to the bottom node is $i_{R} R < 0$ , so the diode is reverse-biased, which confirms it is OFF.
Waveform through R. $i_{R}$ is a half-wave-rectified sinusoid of peak magnitude $I_{m} = π mA$ . It is nonzero only in the negative half-cycles of $i_{s}$ , where it flows upward through R. Because a current source drives the circuit, $i_{R}$ does not depend on the value of R.
Average of a half-wave-rectified sine over one period $T$ , with $ω T = 2 π$ : $\frac{1}{T} \int_{0}^{T /2} I_{m} sin (ω t) d t = \frac{I _{m}}{ω T} [- cos (ω t)]_{0}^{T /2} = \frac{I _{m}}{2 π} \cdot 2 = \frac{I _{m}}{π}$ The same result holds for the negative lobes, with a minus sign: the average of $i_{R}$ is $- \frac{I _{m}}{π}$ .
Substitute: $∣ I_{a v g} ∣ = \frac{π mA}{π} = 1.00 mA$ The minus sign only records that the resistor current flows upward, which is why the question asks for the magnitude. The frequency ( $1500 Hz$ ) does not affect the average.

KEY IDEA

The ideal diode shorts R whenever the source current flows in the diode's forward direction, so R carries only the other half-cycles: a half-wave-rectified sine of peak $π$ mA. Its average over a full period is $I_{m} / π = 1$ mA, whatever the value of R, because a current source drives the circuit.

WATCH OUT FOR

Averaging over only the half-cycle in which R conducts gives $2 I_{m} / π = 2$ mA. The average must be taken over the full period, which gives $I_{m} / π$ . A second trap is to think the answer needs the value of R: with a current source and the diode open, all of $i_{s}$ goes through R, whatever its resistance.

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2023 IN Q.33: solution and answer

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