GATE 2020 IN Q.51: solution and answer

GATE 2020 · INQ.51 · 2 marks · NAT

Assume the diodes in the circuit shown are ideal. The current $I_{x}$ flowing through the $3 k Ω$ resistor (in mA, rounded off to one decimal place) is ____

GATE 2020 IN question 51: A central node X (left) and a node Y (right) are joined by a 3 kΩ resistor; — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

$I_{x} = 1.8 mA$ (only the 15 V and 6 V diodes conduct, and $V_{X} = 9.6 V$ )

Official answer: 1.8Source: Official answer key, GATE 2020 IN (IIT Delhi)

Assumptions. Every diode is ideal: it is a short circuit when it conducts forward current (anode to cathode) and an open circuit otherwise. All terminal voltages are measured from a common ground. Resistances are in $k Ω$ , so a voltage in volts divided by a resistance in $k Ω$ gives a current in $mA$ .
The diode OR on the right. All three right-hand diodes have their anodes on the sources and their cathodes on node Y, so current can only flow from a source into Y. The diode on the highest source conducts: the 15 V diode turns on and $V_{Y} = 15 V$ . The 10 V and 5 V diodes then have their cathodes at 15 V, so they are reverse biased (by 5 V and 10 V) and off.
Which way the left-hand diodes point. Each left-hand diode has its anode on the node-X side and its cathode on its source. So current can only flow out of X, through a $2 k Ω$ resistor, into the 12 V or the 6 V source. That agrees with the marked direction of $I_{x}$ , from Y to X.
First guess: both left-hand diodes ON. KCL at X: $\frac{15 - V _{X}}{3} = \frac{V _{X} - 12}{2} + \frac{V _{X} - 6}{2}$ . Multiplying by 6 gives $30 - 2 V_{X} = 6 V_{X} - 54$ , so $V_{X} = 10.5 V$ .
The check fails. With $V_{X} = 10.5 V < 12 V$ , the 12 V branch would carry $(10.5 - 12) /2 = - 0.75 mA$ , which is against the diode's direction. An ideal diode cannot do that, so the 12 V diode must be OFF.
Second guess: only the 6 V diode ON. KCL at X: $\frac{15 - V _{X}}{3} = \frac{V _{X} - 6}{2}$ , so $30 - 2 V_{X} = 3 V_{X} - 18$ and $V_{X} = \frac{48}{5} = 9.6 V$ .
This state is consistent. No current flows in the 12 V branch, so that diode's anode sits at $V_{X} = 9.6 V$ and its cathode at 12 V. It is reverse biased by 2.4 V and stays OFF. The 6 V diode carries $(9.6 - 6) /2 = 1.8 mA > 0$ , so it is ON.
Current through the $3 k Ω$ resistor: $I_{x} = \frac{V _{Y} - V _{X}}{3 k Ω} = \frac{15 - 9.6}{3} mA = 1.8 mA$ This current is positive, so it flows forward through the 15 V diode, which confirms the assumption in the second step that this diode is ON. It is also equal to the 6 V branch current, as KCL at X requires. Hence $I_{x} = 1.8 mA$ .

KEY IDEA

On the right, the three diodes form an OR circuit that holds node Y at the highest source, 15 V. On the left, guess which diodes are ON, solve KCL at node X, then check every diode: an ON diode must carry forward current, and an OFF diode must be reverse biased. With both left diodes ON, the 12 V branch would need negative current. So only the 6 V diode conducts, $V_{X} = 9.6 V$ and $I_{x} = (15 - 9.6) /3 = 1.8 mA$ .

WATCH OUT FOR

Assuming both left-hand diodes conduct and stopping at $V_{X} = 10.5 V$ , which gives $I_{x} = 1.5 mA$ . That state needs $- 0.75 mA$ through the 12 V diode, which an ideal diode cannot carry. A second mistake is misreading the diode directions: the right-hand diodes pass current INTO node Y (so $V_{Y}$ is set by the highest source, 15 V, not the lowest), and the left-hand diodes pass current OUT of node X.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2020 IN Q.51: solution and answer

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