A capacitor draws an rms current of 0.5 A from a 220 V, 50 Hz ac supply. The same capacitor is now joined to a 110 V, 150 Hz supply. The rms current it now draws is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(0.75\ \mathrm{A}\)
A capacitor limits the current through its capacitive reactance \( X_C=1/2\pi\nu C \). So the current is \( I=2\pi\nu CV \). For the same capacitor, the current is proportional to the product \( \nu V \).
Given:
\( I_1=0.5\ \mathrm{A} \)
\( V_1=220\ \mathrm{V} \)
\( \nu_1=50\ \mathrm{Hz} \)
\( V_2=110\ \mathrm{V} \)
\( \nu_2=150\ \mathrm{Hz} \)
\( \begin{aligned} \frac{I_2}{I_1} &= \frac{\nu_2V_2}{\nu_1V_1} \\ &= \frac{150\times110}{50\times220} \\ &= 1.5 \end{aligned} \)
\( \begin{aligned} I_2 &= 1.5\times I_1 \\ &= 1.5\times0.5 \\ &= 0.75\ \mathrm{A} \end{aligned} \)
Therefore, the current rises to 0.75 A. The frequency becomes three times, while the voltage only becomes half.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “Comparing it to im= vm/R for a purely resistive circuit, we find that (1/ωC) plays the role of resistance. It is called capacitive reactance and …”
\( \displaystyle X_C=\frac{1}{\omega C} \)\( \qquad (7.17) \)
“so that the amplitude of the current is”
\( \displaystyle i_m=\frac{v_m}{X_C} \)\( \qquad (7.18) \)