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.

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Payal Gupta | Contributor-Level 10

8 months ago

7.16 Capacitance of the capacitor, C = 100 $μ F$ = 100 $* 10^{- 6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply voltage, $ν$ = 12 kHz = 12 $* 10^{3}$ Hz

Angular frequency, $ω$ = 2 $π ν$ = 2 $π * 12 * 10^{3}$ rad/s = 24 $π * 10^{3}$ rad/s

For a RC circuit, the impedance Z, is given by Z = $\sqrt[]{R^{2} + \frac{1}{ω^{2} C^{2}}}$

Z = $\sqrt[]{40^{2} + \frac{1}{{(24 π * 10^{3})}^{2} {* (100 * 10^{- 6})}^{2}}}$ = 40 Ω

Peak voltage, $V_{0}$ = $\sqrt 2 * V$ = $\sqrt 2 * 110$ = 155.

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Physics Ncert Solutions Class 12th 2023

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