--- title: "Source current from parallel branch currents" subject: "Basic Electrical Engineering" chapter: 5 key: "semester_1/basic_electrical_engineering:parallel-branch-currents-time-expression" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-branch-currents-time-expression site: BCT Engineering --- # Source current from parallel branch currents Basic Electrical Engineering · Chapter 5: Alternating Quantities - Asked in: 2069 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-branch-currents-time-expression - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-5 ## Question A sinusoidal voltage is applied to three parallel branches yielding branch currents, i1=14.14 Sin(ωt-45°), i2=28.3 Cos(ωt-60°) and i3=7.07 Sin(ωt+60°) (i) Find the complete time expression for the source current (ii) Draw the phasor diagram in terms of effective values. Use the voltage as reference. ## Short answer Convert i2 to sine: 28.3 sin(ωt+30°). Sum phasors (rms): I1=10∠-45°, I2=20∠30°, I3=5∠60° yields I=27.86∠15.1° A rms. Peak source current = 39.4 A. Time expression: i = 39.4 sin(ωt+15.1°) A. ## Answer #### Step 1: Write all currents as sine functions - i₁ = 14.14 sin(ωt − 45°) - i₂ = 28.3 cos(ωt − 60°) = 28.3 sin(ωt − 60° + 90°) = 28.3 sin(ωt + 30°) - i₃ = 7.07 sin(ωt + 60°) #### Step 2: Phasors (rms values, I = Im/√2) with voltage as reference - I₁ = 10.0∠−45° = 7.07 − j7.07 A - I₂ = 20.0∠30° = 17.32 + j10.00 A - I₃ = 5.0∠60° = 2.50 + j4.33 A #### Step 3: Source current I = I₁ + I₂ + I₃ = (7.07 + 17.32 + 2.50) + j(−7.07 + 10.00 + 4.33) = 26.89 + j7.26 A = 27.86∠15.1° A (rms) Peak value: Im = 27.86 × √2 = 39.4 A #### i) Time expression **i = 39.4 sin(ωt + 15.1°) A** (Adding the peak values directly gives the same result: 38.04 + j10.27, magnitude 39.4 A, angle 15.1°.) #### ii) Phasor diagram (effective values, V as reference) ```text I3 = 5 A (60°) / / I2 = 20 A (30°) / / / / I = 27.86 A (15.1°) / / _ - / / _ - ---O-------------------------► V (0°) \ \ \ I1 = 10 A (−45°) ``` I₂ and I₃ lead V, and I₁ lags V by 45°. The resultant I leads V by about 15.1°, so the circuit is net capacitive. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is i2 = 28.3 cos(ωt − 60°) rewritten as a sine so that phasors can be added? A. 28.3 sin(ωt − 150°) B. 28.3 sin(ωt + 150°) C. 28.3 sin(ωt + 30°) D. 28.3 sin(ωt − 30°) Answer: C. cos(x) = sin(x + 90°), so cos(ωt − 60°) = sin(ωt + 30°). (Key point: Convert cosine to sine) 2. Why is the phasor of i1 = 14.14 sin(ωt − 45°) written as 10∠−45° A? A. Peak is used unchanged B. Peak is divided by √2 to get the rms value C. Peak is multiplied by √2 D. Peak is divided by 2 Answer: B. The phasor diagram is in effective values, so I = Im/√2 = 14.14/1.414 = 10 A. (Key point: Rms phasor conversion) 3. How is the source current found from the three branch currents? A. Multiply the three phasors B. Add them as complex numbers (phasor sum) in rectangular form C. Add their magnitudes: 10 + 20 + 5 D. Take the largest branch current Answer: B. In parallel, KCL holds for phasors: I = I1 + I2 + I3 using real and imaginary parts, since the currents have different phase angles. (Key point: Add phasors) 4. The sum is 26.89 + j7.26 A rms. What is the source current expression? A. i = 39.4 sin(ωt + 15.1°) A B. i = 27.86 sin(ωt − 15.1°) A C. i = 27.86 sin(ωt + 15.1°) A D. i = 39.4 sin(ωt − 15.1°) A Answer: A. Convert the rms 27.86 A back to peak (×√2 = 39.4 A) and keep the angle +15.1°, so the current leads the voltage. 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