GATE 2025 EE Q.58: solution and answer

GATE 2025 · EEQ.58 · 2 marks · NAT

In an experiment to measure the active power drawn by a single-phase RL Load connected to an AC source through a $2 Ω$ resistor, three voltmeters are connected as shown in the figure below. The voltmeter readings are as follows: $V_{S o u r ce} = 200 V$ , $V_{R} = 9 V$ , $V_{L o a d} = 199 V$ . Assuming perfect resistors and ideal voltmeters, the Load-active power measured in this experiment, in W, is ________ (round off to one decimal place).

GATE 2025 EE question 58: Single series loop. — Figure from the GATE 2025 EE paper. It is part of the question.

Answer and solution

ANSWER

The load active power measured by the three-voltmeter method is $P = \frac{V _{S}^{2} - V _{R}^{2} - V _{L}^{2}}{2 R} = \frac{318}{4} = 79.5 W$ .

Official answer: 78.0 to 81.0Source: Official answer key, GATE 2025 EE (IIT Roorkee)

This is the three-voltmeter method. A known series resistor $R = 2 Ω$ carries the load current $I$ , and the three voltmeters read the supply voltage $V_{S} = 200 V$ , the resistor voltage $V_{R} = 9 V$ and the load voltage $V_{L} = 199 V$ .
Take the current $I$ as the reference phasor. The resistor voltage $V_{R}$ is in phase with $I$ , and the RL load voltage $V_{L}$ leads $I$ by the load power-factor angle $ϕ$ . The supply voltage is the phasor sum $V_{S} = V_{R} + V_{L}$ , and the angle between $V_{R}$ and $V_{L}$ is $ϕ$ , so by the cosine rule: $V_{S}^{2} = V_{R}^{2} + V_{L}^{2} + 2 V_{R} V_{L} cos ϕ \Rightarrow V_{L} cos ϕ = \frac{V _{S}^{2} - V _{R}^{2} - V _{L}^{2}}{2 V _{R}}$
The load active power is $P = V_{L} I cos ϕ$ with $I = \frac{V _{R}}{R}$ (ideal voltmeters draw no current, so the same $I$ flows in $R$ and the load). Substituting: $P = \frac{V _{R}}{R} \cdot \frac{V _{S}^{2} - V _{R}^{2} - V _{L}^{2}}{2 V _{R}} = \frac{V _{S}^{2} - V _{R}^{2} - V _{L}^{2}}{2 R}$
Substitute the readings: $V_{S}^{2} = 40000$ , $V_{R}^{2} = 81$ , $V_{L}^{2} = 39601$ , so $V_{S}^{2} - V_{R}^{2} - V_{L}^{2} = 40000 - 81 - 39601 = 318 V^{2}$ .
$P = \frac{318}{2 \times 2} = 79.5 W$
Check: $I = 9/2 = 4.5 A$ and $cos ϕ = \frac{318}{2 \times 9 \times 199} = \frac{318}{3582} = 0.0888$ , so $P = 199 \times 4.5 \times 0.0888 = 79.5 W$ . The power factor is about $0.089$ lagging, a highly inductive load, which is consistent with an RL load.

KEY IDEA

Three-voltmeter method: the series resistor voltage is in phase with the current, so the supply voltage is the phasor sum of $V_{R}$ and $V_{L}$ with angle $ϕ$ between them. The cosine rule gives $V_{L} cos ϕ$ , and with $I = V_{R} / R$ the load power reduces to $P = (V_{S}^{2} - V_{R}^{2} - V_{L}^{2}) / (2 R)$ . Only the three readings and $R$ are needed.

WATCH OUT FOR

Adding the voltages arithmetically ( $9 + 199 = 208 \neq = 200$ ) and concluding the data are inconsistent, when they are phasors at an angle. Other mistakes: dividing by $2 V_{R}$ or by $R$ instead of $2 R$ (giving $17.7 W$ or $159 W$ ), including the resistor's own loss $V_{R}^{2} / R = 40.5 W$ in the load power, or using a minus sign in the cosine rule, which gives a negative $cos ϕ$ .

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 8 V, I, power & energy questions, with notes

GATE 2025 EE Q.58: solution and answer

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