Obtain the answers to (a) and (b) in Question 15, if the circuit is connected to 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 d.c. circuit after the steady state.

Solution

(a) Here,
Effective volatge, E = 110 V
Frequency of Ac supply, f = 12 KHz

For the high frequency,

ω = 2 πf = 2 π \times 12 \times 10^{3} rad s^{- 1} .

∴

Maximum current,

I_{0} = \frac{E_{0}}{\sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}} = \frac{2 E_{v}}{\sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}}

\Rightarrow I_{0} = \frac{\sqrt{2} \times 110}{\sqrt{1600 + \frac{1}{4 π^{2} \times 144 \times 10^{6} \times 10^{- 8}}}} A

= \frac{1.414 \times 110}{\sqrt{1600 + 0.0176}} A = \frac{1.414}{40} A = 3.9 A

[It may be noted that the capacitor term is negligible at higher frequencies.]

(b) Now, phase lag

\tan ϕ = \frac{1}{ω C R} = \frac{1}{2 π \times 12 \times 10^{3} \times 10^{- 4} \times 40} = \frac{1}{96 π}

i.e., we can see that, ϕ is nearly zero at high frequency.

It is clear from here that at high frequency, capacitor acts like a conductor.
For a D.C. circuit, after steady state has been reached, ω = 0.
Hence, $χ_{C} = \frac{1}{ωC} = \infty$
Therefore, capacitor C amounts to an open circuit.

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Electrostatic Potential And Capacitance

Obtain the answers to (a) and (b) in Question 15, if the circuit is connected to 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 d.c. circuit after the steady state.

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