A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V,

Solution:

Capacitance of the capacitor, $C=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}$

Resistance of the resistor, $R=40 \Omega$

Supply voltage, V = 110 V

(a) Frequency of oscillations, ν= 60 Hz

Angular frequency, $\omega=2 \pi \nu=2 \pi \times 60 \mathrm{rad} / \mathrm{s}$

For a RC circuit, we have the relation for impedance as:

$Z=R^{2}+\frac{1}{\omega^{2} C^{2}}$

Peak voltage, $V_{0}=V \sqrt{2}=110 \sqrt{2} \mathrm{~V}$

Maximum current is given as:

$I_{0}=\frac{V_{0}}{Z}$

$=\frac{V_{0}}{\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}}$\

$=\frac{110 \sqrt{2}}{\sqrt{(40)^{2}+\frac{1}{(120 \pi)^{2} \times\left(10^{-4}\right)^{2}}}}$

$=\frac{110 \sqrt{2}}{\sqrt{1600+\frac{10^{8}}{(120 \pi)^{2}}}}=3.24 \mathrm{~A}$

(b) In a capacitor circuit, the voltage lags behind the current by a phase angle ofΦ. This angle is given by the relation:

$\therefore \tan \phi=\frac{\frac{1}{\omega C}}{R}=\frac{1}{\omega C R}$

$=\frac{1}{120 \pi \times 10^{-4} \times 40}=0.6635$

$\phi=\tan ^{-1}(0.6635)=33.56^{\circ}$

$=\frac{33.56 \pi}{180} \mathrm{rad}$

$\therefore$ Time lag $=\frac{\phi}{\omega}$

$=\frac{33.56 \pi}{180 \times 120 \pi}=1.55 \times 10^{-3} \mathrm{~s}=1.55 \mathrm{~ms}$

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.