Ncert Exercises & Solutions

7.16 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 behavior with that of a capacitor in a dc circuit after the steady-state.

Given,

Capacitance, 𝐶 = 100 μF

Resistance, 𝑅 = 40 Ω,

rms voltage, 𝑉rms = 110 V

frequency, 𝑓 = 12 kHz

(a) For a capacitive circuit, the maximum current-

$\begin{array}{l} = \frac{\sqrt{2} \times V_{r m s}}{\sqrt{X_{C}^{2} + R^{2}}} \\ = \frac{\sqrt{2} \times 110 V}{\sqrt{(2 π \times 12 k H z \times 100 μ F} Ω)^{2} + (40 Ω)^{2}} = 3.88 A \end{array}$

(b)

$\begin{array}{l} \tan ϕ = \frac{1}{ω C R} = \frac{1}{2 π \times 12 k H 2 \times 100 μ F \times 100 Ω} = \frac{1}{96 π} \\ ϕ = \tan^{- 1} \frac{1}{96 π} = {0.2}^{\circ} = \frac{0.2 π}{180} r a d \\ Timelag = \frac{ϕ}{ω} = \frac{0.2 π}{180 \times 2 π \times 12 \times 10^{3}} = 0.04 μ s \end{array}$

Hence, 𝜙 tends to become 0 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 𝐶𝐶 amounts to an open circuit.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions