Question

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

• A. 2.65A
• B. 2.75A
• C. 2.58A
• D. 2.95A

Answer 1

Answer: (D)

2.95A

Answer 2

: Here, $C = 100\mu F = 100 \times 10^{- 6}F = 10^{- 4}F,$

R = 40 $\Omega$,

$V_{rms} = 100V,\upsilon = 60Hz$

$\therefore V_{0} = \sqrt{2}V_{rms} = 100\sqrt{2}V$

In series RC circuit,

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

$= \frac{1}{\sqrt{R^{2} + \frac{1}{4\pi^{2}\upsilon^{2}C^{2}}}}$ $\lbrack\because\omega = 2\pi\upsilon\rbrack$

Maximum current in the circuit,

$I_{0} = \frac{V_{0}}{Z} = \frac{V_{0}}{\sqrt{R^{2} + \frac{1}{4\pi^{2}\upsilon^{2}C^{2}}}}$

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