Q.10 In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3 Ω , 8 Ω and 4 Ω respectively, calculate phase difference between voltage and current.

Answer: φ = tan⁻¹(4/3) ≈ 53.13^° (about 53^° 8' ), voltage leading the current. tanφ = X_L - X_C/R = 8 - 4/3 = 4/3 = 1.333 φ = tan⁻¹(1.333) ≈ 53.13^° Since X_L gt X_C , the circuit is inductive and the voltage leads the current by about 53^° . (Impedance Z = √(3² + 4²) = 5\ Ω , so cosφ = 0.6 .)

In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3 \(\Omega\), 8 \(\Omega\) and 4 \(\Omega\) respectively, calculate phase difference between voltage and current.

Q.10 In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3 \(\Omega\), 8 \(\Omega\) and 4 \(\Omega\) respectively, calculate phase difference between voltage and current.

Doubtly Asked 28 Oct 2026 Board exam · March 2025 hsc 12th Save WhatsApp Report

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Answer: \(\phi = \tan^{-1}(4/3) \approx 53.13^\circ\) (about \(53^\circ 8'\)), voltage leading the current.

\[\tan\phi = \frac{X_L - X_C}{R} = \frac{8 - 4}{3} = \frac{4}{3} = 1.333\]\[\phi = \tan^{-1}(1.333) \approx 53.13^\circ\]

Since \(X_L \gt X_C\), the circuit is inductive and the voltage leads the current by about \(53^\circ\). (Impedance \(Z = \sqrt{3^2 + 4^2} = 5\ \Omega\), so \(\cos\phi = 0.6\).)

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3 \(\Omega\), 8 \(\Omega\) and 4 \(\Omega\) respectively, calculate phase difference between voltage and current.

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