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Obtain the answers to( a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply.

Q7.16  Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110\: V10kHz  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.

Answers (1)
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Given,

The capacitance of the capacitor C=100\mu F

The resistance of the circuit R=40\Omega

Voltage supply V = 100V

Frequency of voltage supply f=12kHz

The maximum current in the circuit

I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}V}{\sqrt{R^2+\left ( \frac{1}{wC} \right )^2}}=\frac{\sqrt{2}*110}{\sqrt{40^2+\left ( \frac{1}{2\pi *12*10^3*100*10^{-6}} \right )^2}}=3.9A

Hence maximum current in the circuit is 3.9A.

b)

In the case of capacitor, we have 

tan\phi=\frac{\frac{1}{wC}}{R}=\frac{1}{wCR}

So,

tan\phi=\frac{1}{wCR}=\frac{1}{2\pi 10*10^3 *100*10^{-6}*40}=\frac{1}{96\pi}

\phi=0.2^0

So the time lag between max voltage and max current is :

t=\frac{\phi }{w}=\frac{0.2\pi}{180*2\pi*60}=0.04\mu s

At high frequencies, \phi tends to zero. which indicates capacitor acts as a conductor at high frequencies.

In the DC circuit, after a steady state is achieved, Capacitor acts like an open circuit.

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