Question

An alternating emf $E = 110\sqrt{2} \sin 100t \, \text{volt}$ is applied to a capacitor of $2\mu\text{F}$, the rms value of current in the circuit is $\dots$ $\text{mA}$.

Answer 1

Given:
$C = 2 \, \mu\text{F}, \quad E = 110\sqrt{2} \sin(100t).$
The capacitive reactance is:
$X_C = \frac{1}{\omega C},$
where $\omega = 100 \, \text{rad/s}$ and $C = 2 \times 10^{-6} \, \text{F}$.
Substitute:
$X_C = \frac{1}{100 \cdot 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega.$
The peak current is:
$i_0 = \frac{E_0}{X_C},$
where $E_0 = 110\sqrt{2} \, \text{V}$.
Substitute:
$i_0 = \frac{110\sqrt{2}}{5000}.$
The RMS value of current is:$i_{\text{rms}} = \frac{i_0}{\sqrt{2}}.$
Substitute:$i_{\text{rms}} = \frac{110\sqrt{2}}{5000\sqrt{2}} = \frac{110}{5000}.$
Simplify:$i_{\text{rms}} = \frac{110}{5000} \, \text{A} = 22 \, \text{mA}.$
Thus, the RMS value of current in the circuit is:$i_{\text{rms}} = 22 \, \text{mA}.$

Answer 2

Given:
$C = 2 \, \mu\text{F}, \quad E = 110\sqrt{2} \sin(100t).$
The capacitive reactance is:
$X_C = \frac{1}{\omega C},$
where $\omega = 100 \, \text{rad/s}$ and $C = 2 \times 10^{-6} \, \text{F}$.
Substitute:
$X_C = \frac{1}{100 \cdot 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega.$
The peak current is:
$i_0 = \frac{E_0}{X_C},$
where $E_0 = 110\sqrt{2} \, \text{V}$.
Substitute:
$i_0 = \frac{110\sqrt{2}}{5000}.$
The RMS value of current is:$i_{\text{rms}} = \frac{i_0}{\sqrt{2}}.$
Substitute:$i_{\text{rms}} = \frac{110\sqrt{2}}{5000\sqrt{2}} = \frac{110}{5000}.$
Simplify:$i_{\text{rms}} = \frac{110}{5000} \, \text{A} = 22 \, \text{mA}.$
Thus, the RMS value of current in the circuit is:$i_{\text{rms}} = 22 \, \text{mA}.$