--- title: "Parallel RL and RC branch currents, power" subject: "Basic Electrical and Electronics Engineering" chapter: 1 key: "semester_1/basic_electrical_and_electronics_engineering:parallel-impedance-total-current" source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/parallel-impedance-total-current site: BCT Engineering --- # Parallel RL and RC branch currents, power Basic Electrical and Electronics Engineering · Chapter 1: Fundamentals of Electrical and Electronics Circuits - Asked in: 2081 Chaitra (new course), 6 marks - Video: [PARALLEL RL CIRCUITS](https://www.youtube.com/watch?v=3cpRJuvF6f0) — PublicResourceOrg - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/parallel-impedance-total-current - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1 ## Question From the given circuit, find total current (I), branch currents $I_1$ and $I_2$, and real power. [Figure: source $200\angle 53.8°$ V feeding two parallel branches; branch 1 ($I_1$) has $18\ \Omega$ in series with $j24\ \Omega$; branch 2 ($I_2$) has $15\ \Omega$ in series with $-j30\ \Omega$.] ## Short answer Z1 = 18+j24 = 30∠53.13° Ω, Z2 = 15-j30 = 33.54∠-63.43° Ω. I1 = 200∠53.8° / Z1 = 6.667∠0.67° A, I2 = 200∠53.8° / Z2 = 5.963∠117.23° A. Total current I = I1+I2 = 6.667∠53.8° A. Real power = I1²×18 + I2²×15 = 1333.3 W. ## Answer **Given:** $V=200\angle53.8^\circ$ V, $Z_1=18+j24=30\angle53.13^\circ\ \Omega$, $Z_2=15-j30=33.54\angle-63.43^\circ\ \Omega$. **Branch currents:** $$ I_1=\frac{V}{Z_1}=\frac{200\angle53.8^\circ}{30\angle53.13^\circ}=6.667\angle0.67^\circ\ \text{A} $$ $$ I_2=\frac{V}{Z_2}=\frac{200\angle53.8^\circ}{33.54\angle-63.43^\circ}=5.963\angle117.23^\circ\ \text{A} $$ **Total current (phasor sum):** $$ I=I_1+I_2=(6.667+j0.078)+(-2.728+j5.302)=3.939+j5.380 $$ In polar form, $I=6.667\angle53.8^\circ$ A. The total current is in phase with the source voltage: the reactive parts of the two branch admittances cancel ($\frac{1}{Z_1}+\frac{1}{Z_2}=\frac{1}{30}$ S), so the combination behaves as a pure 30 Ω resistance. **Real power:** $$ P=I_1^2R_1+I_2^2R_2=(6.667)^2\times18+(5.963)^2\times15=800+533.3=1333.3\ \text{W} $$ Check with $P=VI\cos\phi=200\times6.667\times\cos0^\circ=1333.3$ W, where $\phi=0$ since $I$ and $V$ have the same angle. **Answer: $I=6.667\angle53.8^\circ$ A, $I_1=6.667\angle0.67^\circ$ A, $I_2=5.963\angle117.23^\circ$ A, $P=1333.3$ W (unity power factor).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is I1 found? A. I1 = V x Z1 B. I1 = V/Z1 using polar division: 200/30 = 6.667 A and angle 53.8° - 53.13° = 0.67° C. I1 = V/|Z1| with angle 53.8° D. I1 = V/(18 + 24) Answer: B. Divide magnitudes and subtract angles: 6.667 angle 0.67° A. (Key point: Branch current) 2. How is the total current found? A. Add magnitudes: 6.667 + 5.963 = 12.63 A B. Take the larger branch current C. Subtract magnitudes D. Add the phasors I1 + I2 (rectangular sum) Answer: D. Currents have different angles, so the phasor sum 3.939 + j5.380 = 6.667 angle 53.8° A is used. (Key point: Total current) 3. Total current has the same angle as the source voltage. What does this imply? A. Unity power factor; reactive currents of the branches cancel B. Purely capacitive circuit C. Purely inductive circuit D. Power factor 0.5 lagging Answer: A. Zero angle between V and I means unity pf, with inductive and capacitive reactive currents cancelling. (Key point: Power factor) 4. Which gives the real power? A. P = I1^2 X1 + I2^2 X2 B. P = V I1 + V I2 C. P = I1^2 R1 + I2^2 R2 = 1333.3 W D. P = 0 because the branches are reactive Answer: C. Only the resistances dissipate power: 800 + 533.3 = 1333.3 W, matching VI cos(0) = 1333.3 W. 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