Question

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

• A. $2.65A$
• B. $2.75A$
• C. $2.85A$
• D. $2.95A$

Answer 1

The correct option is D $2.95A$

$C=100\mu F=100\times {10}^{-6}F$

$R=40\Omega$

$V_{rms}=100v$

$f=60hz$

$\therefore v_0=\sqrt{2}{v}_{rms}=100\sqrt{2}v$

In series $RC$ circuit

$Z=\sqrt{R^2+{X_c}^2}=\sqrt{R^2+\frac{1}{w^2{c}^2}}[w=2\pi f]$

$\therefore$ maximum current in the circuit

$I_0=\frac{v_0}{Z}=\frac{V_0}{\sqrt{R^2+\frac{1}{{4\pi }^2{f}^2{c}^2}}}$

$=\frac{100\sqrt{2}}{\sqrt{{40}^2+\frac{1}{4\times {\left(3.14\right)}^2\times {\left(60\right)}^2\times {\left({10}^{-4}\right)}^2}}}$

$I_o=2.95A$