In the series RLC circuit in the figure, R = 10 Ω, ωL = 30 Ω, 1/(ωC) = 20 Ω and the source voltage is V = 100 V RMS. Which pair correctly gives the RMS value of the current I and its phase relative to the source voltage?

Answer: C. About 7.1 A, lags the voltage by 45°

Method 1 (complex numbers): Z = R + j(ωL − 1/(ωC)) = 10 + j10 Ω, so I = 100/(10 + j10) = 5 − j5 A. Its magnitude is √(5² + 5²) ≈ 7.07 A and its angle is −45°, so the current lags the voltage by 45°. Method 2 (phasor diagram): taking the current as the reference, VR = 10I is in phase, VL = 30I leads by 90° and VC = 20I lags by 90°, so only the difference 10I remains as the reactive part. From |V| = √(10² + 10²) × I ≈ 14.1 × I = 100 V, I ≈ 7.1 A, and the voltage leads the current by 45°, that is, the current lags by 45°. Since ωL > 1/(ωC), the circuit is inductive.

• A. This is calculated with 10 + 30 + 20 = 60 Ω. Reactances are at right angles to resistance, and those of L and C are opposite, so their magnitudes cannot simply be added.
• B. The magnitude is right, but the direction of the phase is reversed. In an inductive circuit with ωL > 1/(ωC), the current lags the voltage.
• C. Correct. I = 5 − j5 A, that is, about 7.1 A lagging by 45°.