Login/Sign Up
Home » 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.
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.
Alternating Current | CBSE | Class 12 | Guides | NCERT | Physics
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, \mathrm{C}=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}

The resistance of the resistor, \mathrm{R}=40 \Omega

Supply voltage, \mathrm{V}=110 \mathrm{~V}

Frequency, v=12 \times 10^{3} \mathrm{~Hz}

Angular frequency, \omega=2 \pi v=2 \pi \times 12 \times 10^{3}=24 \pi \times 10^{3} \mathrm{rad} / \mathrm{s}

Maximum current in the circuit

{{I}_{0}}={{V}_{0}}/Z

{{V}_{0}}=V\sqrt{2}=110\sqrt{2}=155.6~\text{V}

Here, Z=\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}

Z=\sqrt{40^{2}+\frac{1}{\left(24 \pi \times 10^{3}\right)^{2}\left(10^{-4}\right)^{2}}}

Z=\sqrt{1600+\frac{1}{\left(24 \pi \times 10^{3} \times 10^{-4}\right)^{2}}}

Z=\sqrt{1600+0.178}

Z=\sqrt{1600.178}

\text{Z}=40

{{\text{I}}_{0}}=110\sqrt{2}/40

{{\text{I}}_{0}}=155.6/40=3.89~\text{A}

(b) In the capacitor circuit, the voltage will lag behind the current by a phase angle \Phi.

Therefore, \tan \Phi=1 / \omega \mathrm{CR}

=1/24\pi \times {{10}^{3}}\times {{10}^{-4}}\times 40

\tan \Phi =1/96\pi

\Phi =\tan -1(0.0033)={{0.2}^{0}}

=0.2\pi /180\text{rad}

Time lag, \mathrm{t}=\Phi / \omega=0.2 \pi /\left(180 \times 24 \pi \times 10^{3}\right)

=0.2/\left( 4320\times {{10}^{3}} \right)

=4.6\times {{10}^{-8}}~\text{s}=0.04\mu \text{s}

Thus, at high frequency, the capacitor acts like a conductor. For a dc circuit, after steady state, \omega=0, the capacitor amounts to an open circuit.

i

More Material