A circuit element X when connected to an a.c. supply of peak voltage 100 V gives a peak current of 5 A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by π/2. If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

Solution

Physics sir · 4 steps

Read each element from its phase, then add R and X_L as phasors, not as plain numbers.

1. Identify X and Y. Current in phase with voltage means X is a resistor: R = 100/5 = 20 Ω. Current lagging by π/2 means Y is a pure inductor: X_L = 100/5 = 20 Ω.
2. Series impedance. In series, the voltage across R is in phase with the current, and the voltage across L leads it by 90°. So R and X_L add at right angles.

   Z = √(R² + X_L²) = √(20² + 20²) = 20√2 Ω

3. Peak to rms. Peak current I₀ = 100/20√2 = 5/√2 A. The rms value is Iᵣₘₛ = I₀/√2 = 5/2 A.
4. So. The rms current in the series circuit is 5/2 A, option (d).

The idea to keep. The phase of the current tells you the element: in phase is R, lagging by 90° is L, leading by 90° is C. In series, combine them as Z = √(R² + (X_L - X_C)²).

Answer from the NTA final answer key. Solution by PrepWiser.