Question

A 100 $\mu$F capacitor in series with a $40\Omega$ resistor is connected to a 100 V, 60 Hz supply. The time lag between the current maximum and the voltage maximum is

• A. $15.5ms$
• B. $155ms$
• C. $1.55ms$
• D. $1.55s$

Answer 1

The correct option is C $1.55ms$

Capacitance of the capacitor $C=100muF=100\times {10}^{-6}F$

Resistance of the resistor $R=40\Omega$

Supply voltage $V=110V$

Frequency of oscillations $v=60Hz$

Angular frequency $\omega =2\pi v=2\pi \times 60rad/s$

In a capacitor circuit, the voltage lages behind the current by a phase angle of $\varphi$.

This angles is given by the relation:

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

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

$\varphi ={\tan }^{-1}\left(0.6635\right)={33.56}^o$

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

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

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