--- title: "Series AC circuit element parameters" subject: "Basic Electrical and Electronics Engineering" chapter: 1 key: "semester_1/basic_electrical_and_electronics_engineering:series-ac-circuit-parameters" source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/series-ac-circuit-parameters site: BCT Engineering --- # Series AC circuit element parameters Basic Electrical and Electronics Engineering · Chapter 1: Fundamentals of Electrical and Electronics Circuits - Asked in: 2081 Baishakh (new course), 6 marks - Video: [Breaking Down AC Series Circuits Problems](https://www.youtube.com/watch?v=xgDx6ywQfXI) — Pinoy Engineer Online - 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/series-ac-circuit-parameters - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1 ## Question A two-element series circuit is connected across an A.C. source $e = 400\sqrt{2}\sin(\omega t + 40°)$ V. The current in the circuit is then found to be $i = 20\sqrt{2}\cos(314t - 45°)$ A. Determine parameters of the circuit, voltage drop across each element and power factor, power consumed by the circuit. ## Short answer Convert both to sine form: current leads voltage by 5°, so circuit is series R-C. Impedance Z = 400/20 = 20 Ω; R = 20cos5° = 19.92 Ω, XC = 20sin5° = 1.743 Ω, C = 1/(314×1.743) = 1827 μF. Voltage drops: VR = 398.5 V, VC = 34.86 V; power factor = 0.996 leading; power = 7.97 kW. ## Answer **Step 1: Bring both quantities to the same form.** $$ i=20\sqrt2\cos(314t-45^\circ)=20\sqrt2\sin(314t+45^\circ) $$ $$ v=400\sqrt2\sin(314t+40^\circ) $$ RMS values: $V=400$ V, $I=20$ A. Phase angle of current is $+45^\circ$ and of voltage $+40^\circ$, so the current leads the voltage by $$ \phi = 45^\circ-40^\circ = 5^\circ $$ A leading current means the circuit is a series **R–C** circuit. **Step 2: Impedance and parameters.** $$ \begin{aligned} Z &= \frac{V}{I}=\frac{400}{20}=20\ \Omega\\ R &= Z\cos\phi = 20\cos5^\circ = 19.92\ \Omega\\ X_C &= Z\sin\phi = 20\sin5^\circ = 1.743\ \Omega\\ C &= \frac{1}{\omega X_C}=\frac{1}{314\times1.743}=1827\ \mu\text{F} \end{aligned} $$ **Step 3: Voltage drops.** $$ \begin{aligned} V_R &= IR = 20\times19.92 = 398.5\ \text{V}\\ V_C &= IX_C = 20\times1.743 = 34.86\ \text{V} \end{aligned} $$ Check: $\sqrt{398.5^2+34.86^2}=400$ V. **Step 4: Power factor and power.** $$ \cos\phi=\cos5^\circ=0.996\ \text{(leading)} $$ $$ P = VI\cos\phi = 400\times20\times0.996 = 7970\ \text{W}\ (=I^2R) $$ **Answer: $R=19.92\ \Omega$, $C=1827\ \mu$F ($X_C=1.743\ \Omega$); $V_R=398.5$ V, $V_C=34.86$ V; pf = 0.996 leading; $P=7.97$ kW.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is i = 20 sqrt(2) cos(314t - 45°) compared with the sine voltage? A. Rewrite it as 20 sqrt(2) sin(314t + 45°) B. Rewrite it as 20 sqrt(2) sin(314t - 45°) C. Rewrite it as 20 sqrt(2) sin(314t - 135°) D. Use it unchanged since cos and sin are the same Answer: A. cos(x) = sin(x + 90°), so cos(314t - 45°) = sin(314t + 45°). (Key point: Same-form conversion) 2. Current is at +45° and voltage at +40°. What kind of circuit is it? A. Series R-L, current lags by 5° B. Series R-C, current leads by 5° C. Series R-L, current leads by 5° D. Purely resistive Answer: B. The current leads the voltage by 5°, which indicates a capacitive (R-C) circuit. (Key point: Circuit type) 3. With Z = 400/20 = 20 Ω and phase angle 5°, which gives R and XC? A. R = Z sin5° = 1.743 Ω, XC = Z cos5° = 19.92 Ω B. R = Z cos5° = 19.92 Ω, XC = Z sin5° = 1.743 Ω C. R = Z = 20 Ω, XC = 0 D. R = Z tan5° and XC = Z / tan5° Answer: B. The impedance triangle gives R = Z cos(phi) and X = Z sin(phi). (Key point: Parameters R and XC) 4. What are the power factor and power consumed? A. 0.996 lagging, 7.97 kW B. 0.087 leading, 0.7 kW C. 0.996 leading, 7.97 kW D. 1.0, 8 kW with V_C = 0 Answer: C. cos5° = 0.996 leading, and P = VI cos(phi) = 400 x 20 x 0.996 = 7970 W (= I^2 R). 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