Answer to Question #130553 in Electric Circuits for Sumedh

Question

For a circuit with 100 microfarad capacitor in series with 25 ohm resistor, what would be the approximate voltage drop across the capacitor if the series voltage was 45 volts and 60 Hz?

Answer 1

As per the question,

Capacitance C=100$\mu$ f=10$^{-4}$ f

Resistance R=25 $\Omega$

Voltage V=45 volts

frequency f=60Hz

Capacitive reactance X$_c=\frac{1}{\omega C}$

=$\frac{1}{2\Pi f\times C}$

=$\frac{1}{2\times 3.14 \times 60 \times 10^{-4}}$

=26.53$\Omega$

Impedence Z=$\sqrt{R^2+X_c^2}$

=$\sqrt{25^2+26.53^2}$

on solving we get

Z=36.45$\Omega$

Current in the circuit

i=$\frac{voltage}{impedence}$

=$\frac{45}{36.53}$

i=1.2345 Ampere

Voltage drop across resistor

say V$_1$ = iR

=1.2345$\times$ 25

=30.86 volts

So Voltage drop across capacitor

= Total voltage - voltage across resistor

=45-30.86

14.14 volts

Hence Voltage drop across capacitor is 14.14 volts.