PYQs / GATE EE / 2021 / Set 1 / Q49 GATE EE 2021 Set 1 — Question 49 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 Medium Unbalanced Circuits AC Power & Three-Phase Electric Circuits
Electric Circuits → AC Power & Three-Phase → Unbalanced Circuits
Last updated 5 September 2026
Question A three-phase balanced voltage is applied to the load shown. The phase sequence is RYB. The ratio
∣ I B ∣ ∣ I R ∣ \frac{|I_B|}{|I_R|} ∣ I R ∣ ∣ I B ∣ is
________ .
Solution Circuit.
The capacitor
− j 10 Ω -j10\,\Omega − j 10 Ω joins line
R to a common node
N N N ; the inductor
j 10 Ω j10\,\Omega j 10 Ω joins node
N N N to line
Y ; line
B is connected straight to node
N N N . Hence
V N = V B V_N = V_B V N = V B .
Supply phasors (sequence RYB). V R = V ∠ 0 ∘ , V Y = V ∠ − 120 ∘ , V B = V ∠ + 120 ∘ V_R = V\angle 0^\circ,\qquad V_Y = V\angle{-120^\circ},\qquad V_B = V\angle{+120^\circ} V R = V ∠ 0 ∘ , V Y = V ∠ − 12 0 ∘ , V B = V ∠ + 12 0 ∘ Current in the capacitor branch (I R I_R I R V R B = V ∠ 0 ∘ − V ∠ 120 ∘ = 3 V ∠ − 30 ∘ V_{RB} = V\angle 0^\circ - V\angle 120^\circ = \sqrt{3}\,V\angle{-30^\circ} V R B = V ∠ 0 ∘ − V ∠12 0 ∘ = 3 V ∠ − 3 0 ∘ I R = V R B − j 10 = 3 V ∠ − 30 ∘ 10 ∠ − 90 ∘ = 3 V 10 ∠ 60 ∘ I_R = \frac{V_{RB}}{-j10} = \frac{\sqrt{3}\,V\angle{-30^\circ}}{10\angle{-90^\circ}} = \frac{\sqrt{3}\,V}{10}\angle 60^\circ I R = − j 10 V R B = 10∠ − 9 0 ∘ 3 V ∠ − 3 0 ∘ = 10 3 V ∠6 0 ∘ Current in the inductor branch (I Y I_Y I Y V Y B = V ∠ − 120 ∘ − V ∠ 120 ∘ = 3 V ∠ − 90 ∘ V_{YB} = V\angle{-120^\circ} - V\angle 120^\circ = \sqrt{3}\,V\angle{-90^\circ} V Y B = V ∠ − 12 0 ∘ − V ∠12 0 ∘ = 3 V ∠ − 9 0 ∘ I Y = V Y B j 10 = 3 V ∠ − 90 ∘ 10 ∠ 90 ∘ = 3 V 10 ∠ − 180 ∘ I_Y = \frac{V_{YB}}{j10} = \frac{\sqrt{3}\,V\angle{-90^\circ}}{10\angle 90^\circ} = \frac{\sqrt{3}\,V}{10}\angle{-180^\circ} I Y = j 10 V Y B = 10∠9 0 ∘ 3 V ∠ − 9 0 ∘ = 10 3 V ∠ − 18 0 ∘ KCL at node N N N
All three line currents are shown entering the load, so
I R + I Y + I B = 0 ⇒ I B = − ( I R + I Y ) I_R + I_Y + I_B = 0 \quad\Rightarrow\quad I_B = -(I_R + I_Y) I R + I Y + I B = 0 ⇒ I B = − ( I R + I Y ) I R + I Y = 3 V 10 [ ( 0.5 + j 0.866 ) + ( − 1 ) ] = 3 V 10 ( − 0.5 + j 0.866 ) = 3 V 10 ∠ 120 ∘ I_R + I_Y = \frac{\sqrt{3}\,V}{10}\left[(0.5 + j0.866) + (-1)\right] = \frac{\sqrt{3}\,V}{10}\,(-0.5 + j0.866) = \frac{\sqrt{3}\,V}{10}\angle 120^\circ I R + I Y = 10 3 V [ ( 0.5 + j 0.866 ) + ( − 1 ) ] = 10 3 V ( − 0.5 + j 0.866 ) = 10 3 V ∠12 0 ∘ I B = 3 V 10 ∠ − 60 ∘ I_B = \frac{\sqrt{3}\,V}{10}\angle{-60^\circ} I B = 10 3 V ∠ − 6 0 ∘ Ratio of magnitudes. ∣ I R ∣ = 3 V 10 , ∣ I B ∣ = 3 V 10 |I_R| = \frac{\sqrt{3}\,V}{10},\qquad |I_B| = \frac{\sqrt{3}\,V}{10} ∣ I R ∣ = 10 3 V , ∣ I B ∣ = 10 3 V ∣ I B ∣ ∣ I R ∣ = 1 \frac{|I_B|}{|I_R|} = 1 ∣ I R ∣ ∣ I B ∣ = 1 Answer: 1 1 1 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.
Solution Circuit.
The capacitor
− j 10 Ω -j10\,\Omega − j 10 Ω joins line
R to a common node
N N N ; the inductor
j 10 Ω j10\,\Omega j 10 Ω joins node
N N N to line
Y ; line
B is connected straight to node
N N N . Hence
V N = V B V_N = V_B V N = V B .
Supply phasors (sequence RYB). V R = V ∠ 0 ∘ , V Y = V ∠ − 120 ∘ , V B = V ∠ + 120 ∘ V_R = V\angle 0^\circ,\qquad V_Y = V\angle{-120^\circ},\qquad V_B = V\angle{+120^\circ} V R = V ∠ 0 ∘ , V Y = V ∠ − 12 0 ∘ , V B = V ∠ + 12 0 ∘ Current in the capacitor branch (I R I_R I R V R B = V ∠ 0 ∘ − V ∠ 120 ∘ = 3 V ∠ − 30 ∘ V_{RB} = V\angle 0^\circ - V\angle 120^\circ = \sqrt{3}\,V\angle{-30^\circ} V R B = V ∠ 0 ∘ − V ∠12 0 ∘ = 3 V ∠ − 3 0 ∘ I R = V R B − j 10 = 3 V ∠ − 30 ∘ 10 ∠ − 90 ∘ = 3 V 10 ∠ 60 ∘ I_R = \frac{V_{RB}}{-j10} = \frac{\sqrt{3}\,V\angle{-30^\circ}}{10\angle{-90^\circ}} = \frac{\sqrt{3}\,V}{10}\angle 60^\circ I R = − j 10 V R B = 10∠ − 9 0 ∘ 3 V ∠ − 3 0 ∘ = 10 3 V ∠6 0 ∘ Current in the inductor branch (I Y I_Y I Y V Y B = V ∠ − 120 ∘ − V ∠ 120 ∘ = 3 V ∠ − 90 ∘ V_{YB} = V\angle{-120^\circ} - V\angle 120^\circ = \sqrt{3}\,V\angle{-90^\circ} V Y B = V ∠ − 12 0 ∘ − V ∠12 0 ∘ = 3 V ∠ − 9 0 ∘ I Y = V Y B j 10 = 3 V ∠ − 90 ∘ 10 ∠ 90 ∘ = 3 V 10 ∠ − 180 ∘ I_Y = \frac{V_{YB}}{j10} = \frac{\sqrt{3}\,V\angle{-90^\circ}}{10\angle 90^\circ} = \frac{\sqrt{3}\,V}{10}\angle{-180^\circ} I Y = j 10 V Y B = 10∠9 0 ∘ 3 V ∠ − 9 0 ∘ = 10 3 V ∠ − 18 0 ∘ KCL at node N N N
All three line currents are shown entering the load, so
I R + I Y + I B = 0 ⇒ I B = − ( I R + I Y ) I_R + I_Y + I_B = 0 \quad\Rightarrow\quad I_B = -(I_R + I_Y) I R + I Y + I B = 0 ⇒ I B = − ( I R + I Y ) I R + I Y = 3 V 10 [ ( 0.5 + j 0.866 ) + ( − 1 ) ] = 3 V 10 ( − 0.5 + j 0.866 ) = 3 V 10 ∠ 120 ∘ I_R + I_Y = \frac{\sqrt{3}\,V}{10}\left[(0.5 + j0.866) + (-1)\right] = \frac{\sqrt{3}\,V}{10}\,(-0.5 + j0.866) = \frac{\sqrt{3}\,V}{10}\angle 120^\circ I R + I Y = 10 3 V [ ( 0.5 + j 0.866 ) + ( − 1 ) ] = 10 3 V ( − 0.5 + j 0.866 ) = 10 3 V ∠12 0 ∘ I B = 3 V 10 ∠ − 60 ∘ I_B = \frac{\sqrt{3}\,V}{10}\angle{-60^\circ} I B = 10 3 V ∠ − 6 0 ∘ Ratio of magnitudes. ∣ I R ∣ = 3 V 10 , ∣ I B ∣ = 3 V 10 |I_R| = \frac{\sqrt{3}\,V}{10},\qquad |I_B| = \frac{\sqrt{3}\,V}{10} ∣ I R ∣ = 10 3 V , ∣ I B ∣ = 10 3 V ∣ I B ∣ ∣ I R ∣ = 1 \frac{|I_B|}{|I_R|} = 1 ∣ I R ∣ ∣ I B ∣ = 1 Answer: 1 1 1 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 ) ← Q48 Full paper Q50 → 
