Question

(1) A $100\mu F$ capacitor in series with a $40\Omega$ resistance is connected to a $110V,60Hz$ supply.
What is the maximum current in the circuit?

(2) A $100\mu F$ capacitor in series with a $40\Omega$ resistance is connected to a $110V,12Hz$ supply.
What is the time lag between the current maximum and the voltage maximum?
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.

Answer 1

(1)Step 1: Find circuit impedance

Given,$V_{rms}=110V,f=12Hz$
$C=100\mu F,R=40\Omega$

The circuit impedance is given by $Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}$


$Z=\sqrt{(40{)}^2+{\left(\frac{1}{2\pi \times 12\times {10}^3\times 100\times {10}^{-6}}\right)}^2}$

$Z=\sqrt{(40{)}^2+(0.133{)}^2}$

$Z=40\Omega$

Step 2: Find maximum current

For an RC circuit if, $V=V_0\sin \omega t$

In RC circuit, current $\left(I\right)$ lags from voltage $\left(V\right)$.

$I=I_0\sin (\omega t+\varphi )$

$I=\frac{V_0}{Z}\sin (\omega t+\varphi )$

Then the maximum current is,

$I_0=\frac{V_0}{Z}$ ....(i)

Where, $V_0=\sqrt{2}{V}_{rms}=110\sqrt{2}V$

Hence from equation (i),

$I_0=\frac{110\sqrt{2}}{40}=3.24A$

$I_0=3.88A$

Final Answer: $3.88A$

(2)Step 1: Calculate phase angle

Given,$V_{rms}=110V,f=12Hz$
$C=100\mu F,R=40\Omega$
Inductive reactance,

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

$=\frac{1}{2\pi \times 12\times {10}^3\times 100\times {10}^{-6}}$

$=0.133\Omega$

Now, from phasor diagram

$\tan \varphi =\frac{X_C}{R}$

$\varphi ={\tan }^{-1}\left(\frac{-0.133}{40}\right)$

$\varphi =-{0.2}^{\circ }$

Step 2: Find the time lag between the current maximum and the voltage maximum

For an RC circuit let, $V=V_0\cos \omega t$

Voltage is maximum when $t$ is zero.

Current in a RC circuit, $I=I_0\cos (\omega t+\varphi )$

Current is maximum when

$\omega t+\varphi =0$
$\Rightarrow =t=-\frac{\varphi }{\omega }$ ....(i)

$t=\frac{-{0.2}^{\circ }\times \frac{\pi }{{180}^{\circ }}}{2\pi \times 12\times {10}^3}$

$t=0.04\times {10}^{-6}s=0.04\mu s$

Step 3: Explanation of statement

As $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$ is very low $\left(0.133\Omega \right)$ at high frequency, hence capacitor behaves as conductor (short circuit).

For a dc circuit, after steady state, $\omega =0$ and $C$ amounts to an open circuit.

Final Answer: $0.04\mu s$

As is very low $\left(0.133\Omega \right)$ at high frequency, capacitor behaves as conductor (short circuit).

For a dc circuit, after steady state, $\omega =0$ and $C$ amounts to an open circuit.