Question

The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 175 V and frequency 61.0 Hz applied across the capacitor is to produce a current amplitude of 0.849 A through the capacitor. You may want to review (Pages 1023 - 1028) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A resistor and a capacitor in an ac circuit.

Answer 1

To solve this exercise we will need the concepts concerning impedance and capacitive reactance.

The potential in terms of impedance is given by,

$V = I*\xi$

Where,

$I = current\\\xi = Impedance$

Impedance is equal to

$\xi = \frac{V}{I}$

$\xi = \frac{175}{0.849}$

$\xi = 206.12\Omega$

For definition we know that Impedance is equal also to

$\xi = \frac{1}{Wc} = \frac{1}{2\pi f*c}$

Where f is the frequency and c the capacitive reactance.

Re-arrange for c, we have,

$c = \frac{1}{2\pi f*\xi}$

$c = \frac{1}{2\pi 61*206.12}$

$c = 1.2658*10^{-5} F$