PYQs / GATE EE / 2024 / Set 1 / Q50 GATE EE 2024 Set 1 — Question 50 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Hard Two-Wattmeter Method AC Power & Three-Phase Electric Circuits Unbalanced Circuits
Electric Circuits → AC Power & Three-Phase → Two-Wattmeter Method
Last updated 5 September 2026
Question In the circuit shown,
Z 1 = 50 ∠ − 90 ∘ Ω Z_1 = 50\angle - 90^\circ \Omega Z 1 = 50∠ − 9 0 ∘ Ω and
Z 2 = 200 ∠ − 30 ∘ Ω Z_2 = 200\angle - 30^\circ \Omega Z 2 = 200∠ − 3 0 ∘ Ω . It is supplied by a three phase
400 V 400 \text{ V} 400 V source with the phase sequence being R-Y-B. Assume the watt meters
W 1 W_1 W 1 and
W 2 W_2 W 2 to be ideal. The magnitude of the difference between the readings of
W 1 W_1 W 1 and
W 2 W_2 W 2 in watts is
___________ (rounded off to 2 decimal places).
Correct answer 692.8 to 692.84
Solution Reading the connection from the figure
Z 1 Z_1 Z 1 R and B ; Z 2 Z_2 Z 2 R and Y .W 1 W_1 W 1 + + + W 1 = R e { V R B I R ∗ } W_1 = \mathrm{Re}\{V_{RB} I_R^*\} W 1 = Re { V R B I R ∗ } W 2 W_2 W 2 + + + W 2 = R e { V Y B I Y ∗ } W_2 = \mathrm{Re}\{V_{YB} I_Y^*\} W 2 = Re { V Y B I Y ∗ }
Line voltages (400 V, sequence R-Y-B). Taking
V R Y = 400 ∠ 0 ∘ V_{RY} = 400\angle 0^\circ V R Y = 400∠ 0 ∘ V:
V Y B = 400 ∠ − 120 ∘ V , V B R = 400 ∠ 120 ∘ V , V R B = − V B R = 400 ∠ − 60 ∘ V V_{YB} = 400\angle -120^\circ\ \text{V}, \qquad V_{BR} = 400\angle 120^\circ\ \text{V}, \qquad V_{RB} = -V_{BR} = 400\angle -60^\circ\ \text{V} V Y B = 400∠ − 12 0 ∘ V , V B R = 400∠12 0 ∘ V , V R B = − V B R = 400∠ − 6 0 ∘ V Branch currents I Z 1 = V R B Z 1 = 400 ∠ − 60 ∘ 50 ∠ − 90 ∘ = 8 ∠ 30 ∘ A ( R → B ) I_{Z_1} = \frac{V_{RB}}{Z_1} = \frac{400\angle -60^\circ}{50\angle -90^\circ} = 8\angle 30^\circ\ \text{A} \quad (\text{R} \to \text{B}) I Z 1 = Z 1 V R B = 50∠ − 9 0 ∘ 400∠ − 6 0 ∘ = 8∠3 0 ∘ A ( R → B ) I Z 2 = V R Y Z 2 = 400 ∠ 0 ∘ 200 ∠ − 30 ∘ = 2 ∠ 30 ∘ A ( R → Y ) I_{Z_2} = \frac{V_{RY}}{Z_2} = \frac{400\angle 0^\circ}{200\angle -30^\circ} = 2\angle 30^\circ\ \text{A} \quad (\text{R} \to \text{Y}) I Z 2 = Z 2 V R Y = 200∠ − 3 0 ∘ 400∠ 0 ∘ = 2∠3 0 ∘ A ( R → Y ) Line currents into the load I R = I Z 1 + I Z 2 = 8 ∠ 30 ∘ + 2 ∠ 30 ∘ = 10 ∠ 30 ∘ A I_R = I_{Z_1} + I_{Z_2} = 8\angle 30^\circ + 2\angle 30^\circ = 10\angle 30^\circ\ \text{A} I R = I Z 1 + I Z 2 = 8∠3 0 ∘ + 2∠3 0 ∘ = 10∠3 0 ∘ A I Y = − I Z 2 = 2 ∠ − 150 ∘ A I_Y = -I_{Z_2} = 2\angle -150^\circ\ \text{A} I Y = − I Z 2 = 2∠ − 15 0 ∘ A Wattmeter readings W 1 = R e { V R B I R ∗ } = R e { 400 ∠ − 60 ∘ ⋅ 10 ∠ − 30 ∘ } = R e { 4000 ∠ − 90 ∘ } = 0 W W_1 = \mathrm{Re}\{V_{RB} I_R^*\} = \mathrm{Re}\{400\angle -60^\circ \cdot 10\angle -30^\circ\} = \mathrm{Re}\{4000\angle -90^\circ\} = 0\ \text{W} W 1 = Re { V R B I R ∗ } = Re { 400∠ − 6 0 ∘ ⋅ 10∠ − 3 0 ∘ } = Re { 4000∠ − 9 0 ∘ } = 0 W W 2 = R e { V Y B I Y ∗ } = R e { 400 ∠ − 120 ∘ ⋅ 2 ∠ 150 ∘ } = R e { 800 ∠ 30 ∘ } = 800 cos 30 ∘ = 692.82 W W_2 = \mathrm{Re}\{V_{YB} I_Y^*\} = \mathrm{Re}\{400\angle -120^\circ \cdot 2\angle 150^\circ\} = \mathrm{Re}\{800\angle 30^\circ\} = 800\cos 30^\circ = 692.82\ \text{W} W 2 = Re { V Y B I Y ∗ } = Re { 400∠ − 12 0 ∘ ⋅ 2∠15 0 ∘ } = Re { 800∠3 0 ∘ } = 800 cos 3 0 ∘ = 692.82 W Check: Z 1 Z_1 Z 1 is purely capacitive, so it absorbs no average power; the total power is that of
Z 2 Z_2 Z 2 alone,
400 2 200 cos 30 ∘ = 692.82 \dfrac{400^2}{200}\cos 30^\circ = 692.82 200 40 0 2 cos 3 0 ∘ = 692.82 W
= W 1 + W 2 = W_1 + W_2 = W 1 + W 2 .
∣ W 1 − W 2 ∣ = ∣ 0 − 692.82 ∣ = 692.82 W |W_1 - W_2| = |0 - 692.82| = 692.82\ \text{W} ∣ W 1 − W 2 ∣ = ∣0 − 692.82∣ = 692.82 W Answer: 692.82 692.82 692.82 W
Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer 692.8 to 692.84
Solution Reading the connection from the figure
Z 1 Z_1 Z 1 R and B ; Z 2 Z_2 Z 2 R and Y .W 1 W_1 W 1 + + + W 1 = R e { V R B I R ∗ } W_1 = \mathrm{Re}\{V_{RB} I_R^*\} W 1 = Re { V R B I R ∗ } W 2 W_2 W 2 + + + W 2 = R e { V Y B I Y ∗ } W_2 = \mathrm{Re}\{V_{YB} I_Y^*\} W 2 = Re { V Y B I Y ∗ }
Line voltages (400 V, sequence R-Y-B). Taking
V R Y = 400 ∠ 0 ∘ V_{RY} = 400\angle 0^\circ V R Y = 400∠ 0 ∘ V:
V Y B = 400 ∠ − 120 ∘ V , V B R = 400 ∠ 120 ∘ V , V R B = − V B R = 400 ∠ − 60 ∘ V V_{YB} = 400\angle -120^\circ\ \text{V}, \qquad V_{BR} = 400\angle 120^\circ\ \text{V}, \qquad V_{RB} = -V_{BR} = 400\angle -60^\circ\ \text{V} V Y B = 400∠ − 12 0 ∘ V , V B R = 400∠12 0 ∘ V , V R B = − V B R = 400∠ − 6 0 ∘ V Branch currents I Z 1 = V R B Z 1 = 400 ∠ − 60 ∘ 50 ∠ − 90 ∘ = 8 ∠ 30 ∘ A ( R → B ) I_{Z_1} = \frac{V_{RB}}{Z_1} = \frac{400\angle -60^\circ}{50\angle -90^\circ} = 8\angle 30^\circ\ \text{A} \quad (\text{R} \to \text{B}) I Z 1 = Z 1 V R B = 50∠ − 9 0 ∘ 400∠ − 6 0 ∘ = 8∠3 0 ∘ A ( R → B ) I Z 2 = V R Y Z 2 = 400 ∠ 0 ∘ 200 ∠ − 30 ∘ = 2 ∠ 30 ∘ A ( R → Y ) I_{Z_2} = \frac{V_{RY}}{Z_2} = \frac{400\angle 0^\circ}{200\angle -30^\circ} = 2\angle 30^\circ\ \text{A} \quad (\text{R} \to \text{Y}) I Z 2 = Z 2 V R Y = 200∠ − 3 0 ∘ 400∠ 0 ∘ = 2∠3 0 ∘ A ( R → Y ) Line currents into the load I R = I Z 1 + I Z 2 = 8 ∠ 30 ∘ + 2 ∠ 30 ∘ = 10 ∠ 30 ∘ A I_R = I_{Z_1} + I_{Z_2} = 8\angle 30^\circ + 2\angle 30^\circ = 10\angle 30^\circ\ \text{A} I R = I Z 1 + I Z 2 = 8∠3 0 ∘ + 2∠3 0 ∘ = 10∠3 0 ∘ A I Y = − I Z 2 = 2 ∠ − 150 ∘ A I_Y = -I_{Z_2} = 2\angle -150^\circ\ \text{A} I Y = − I Z 2 = 2∠ − 15 0 ∘ A Wattmeter readings W 1 = R e { V R B I R ∗ } = R e { 400 ∠ − 60 ∘ ⋅ 10 ∠ − 30 ∘ } = R e { 4000 ∠ − 90 ∘ } = 0 W W_1 = \mathrm{Re}\{V_{RB} I_R^*\} = \mathrm{Re}\{400\angle -60^\circ \cdot 10\angle -30^\circ\} = \mathrm{Re}\{4000\angle -90^\circ\} = 0\ \text{W} W 1 = Re { V R B I R ∗ } = Re { 400∠ − 6 0 ∘ ⋅ 10∠ − 3 0 ∘ } = Re { 4000∠ − 9 0 ∘ } = 0 W W 2 = R e { V Y B I Y ∗ } = R e { 400 ∠ − 120 ∘ ⋅ 2 ∠ 150 ∘ } = R e { 800 ∠ 30 ∘ } = 800 cos 30 ∘ = 692.82 W W_2 = \mathrm{Re}\{V_{YB} I_Y^*\} = \mathrm{Re}\{400\angle -120^\circ \cdot 2\angle 150^\circ\} = \mathrm{Re}\{800\angle 30^\circ\} = 800\cos 30^\circ = 692.82\ \text{W} W 2 = Re { V Y B I Y ∗ } = Re { 400∠ − 12 0 ∘ ⋅ 2∠15 0 ∘ } = Re { 800∠3 0 ∘ } = 800 cos 3 0 ∘ = 692.82 W Check: Z 1 Z_1 Z 1 is purely capacitive, so it absorbs no average power; the total power is that of
Z 2 Z_2 Z 2 alone,
400 2 200 cos 30 ∘ = 692.82 \dfrac{400^2}{200}\cos 30^\circ = 692.82 200 40 0 2 cos 3 0 ∘ = 692.82 W
= W 1 + W 2 = W_1 + W_2 = W 1 + W 2 .
∣ W 1 − W 2 ∣ = ∣ 0 − 692.82 ∣ = 692.82 W |W_1 - W_2| = |0 - 692.82| = 692.82\ \text{W} ∣ W 1 − W 2 ∣ = ∣0 − 692.82∣ = 692.82 W Answer: 692.82 692.82 692.82 W
Understand the concept, then try another question Revisit Electric Circuits with concept notes, common mistakes and an original worked example before your next attempt.
More questions on AC Power & Three-Phase 2026 Set 1 Q23 A circuit with ideal elements is shown. [figure] Which one of the following options correctly… 2026 Set 1 Q24 For the circuit shown, which one of the following options correctly identifies the Thevenin’s… 2026 Set 1 Q25 Refer to the four circuits shown. Which one of the following options for k 1 , k 2 , k 3 , k_1, k_2, k_3, k 1 , k 2 , k 3 , k 4 k_4 k 4 2026 Set 1 Q26 A single-phase voltage source v s = 325 sin ( 2 π 50 t ) v_s = 325 \sin(2\pi 50t) v s = 325 sin ( 2 π 50 t ) 2026 Set 1 Q42 An electrical component has voltage drop v = V m sin ( ω t ) v = V_m \sin(\omega t) v = V m sin ( ω t ) ← Q49 Full paper Q51 → 
