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 = (V)/(I)`

`\xi = (175)/(0.849)`

`\xi = 206.12\Omega`

For definition we know that Impedance is equal also to

`\xi = (1)/(Wc) = (1)/(2\pi f*c)`

Where f is the frequency and c the capacitive reactance.

Re-arrange for c, we have,

`c = (1)/(2\pi f*\xi)`

`c = (1)/(2\pi 61*206.12)`

`c = 1.2658*10^(-5) F`