Obtain the answers to


Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.


Capacitance of the capacitor, C = 100 μF = 100 × 10−6 F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply, ν = 12 kHz = 12 × 103 Hz

Angular Frequency, ω = 2 πν= 2 × π × 12 × 10303

= 24π × 103 rad/s

Peak voltage, $V_{0}=V \sqrt{2}=110 \sqrt{2} \mathrm{~V}$

Maximum current, $I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}}$

$=\frac{110 \sqrt{2}}{\sqrt{(40)^{2}+\frac{1}{\left(24 \pi \times 10^{3} \times 100 \times 10^{-6}\right)^{2}}}}$

$=\frac{110 \sqrt{2}}{\sqrt{1600+\left(\frac{10}{24 \pi}\right)^{2}}}=3.9 \mathrm{~A}$

For an RC circuit, the voltage lags behind the current by a phase angle of Φ given as:

$\tan \phi=\frac{\frac{1}{\omega C}}{R}=\frac{1}{\omega C R}$

$=\frac{1}{24 \pi \times 10^{3} \times 100 \times 10^{-6} \times 40}$

$\tan \phi=\frac{1}{96 \pi}$

$\therefore \phi \simeq 0.2^{\circ}$

$=\frac{0.2 \pi}{180} \mathrm{rad}$

$\therefore$ Time lag $=\frac{\phi}{\omega}$

$=\frac{0.2 \pi}{180 \times 24 \pi \times 10^{3}}=1.55 \times 10^{-3} \mathrm{~s}=0.04 \mu \mathrm{s}$

Hence, Φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.

In a dc circuit, after the steady state is achieved, ω = 0. Hence, capacitor C amounts to an open circuit.

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