A resistor of 30 Ω, an inductor of reactance 80 Ω and a capacitor of reactance 40 Ω are joined in series across a 200 V (rms) ac source. The rms voltage across the inductor and the capacitor taken together is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(160\ \mathrm{V}\)
The same current flows through all three elements. The voltages across the inductor and the capacitor are opposite in phase. So they subtract, and the voltage across the pair is \( I\,|X_L-X_C| \).
\( \begin{aligned} Z &= \sqrt{R^{2}+(X_C-X_L)^{2}} \\ &= \sqrt{30^{2}+(40-80)^{2}} \\ &= \sqrt{2500} \\ &= 50\ \Omega \end{aligned} \)
\( \begin{aligned} I &= \frac{V}{Z} \\ &= \frac{200}{50} \\ &= 4\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} V_{LC} &= I\,|X_L-X_C| \\ &= 4\times(80-40) \\ &= 160\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Phasor-diagram solution – “… Since VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude |vCm – vLm|. …”
Phasor-diagram solution – “By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit:”
\( \displaystyle i_m=\frac{v_m}{Z} \qquad [7.25(b)] \)
“where”
\( \displaystyle Z=\sqrt{R^{2}+(X_C-X_L)^{2}} \qquad (7.26) \)