Question

An ac generator with a frequency of 30.0 Hz and an rms voltage of 14.0 V is connected to a 65.0 μF capacitor.

1. What is the maximum current in this circuit?

2. What is the current in the circuit when the voltage across the capacitor is 4.30 V and increasing?

3. What is the current in the circuit when the voltage across the capacitor is 4.30 V and decreasing?

Answer 1

Let the frequency of the ac be f = 30.0 Hz

Let the rms voltage be V_rms = 14.0 V

Let the capacitance be C = 65 F

a) To Calculate the maximum current , we need to calculate the capacitive reactance X_c

X_c = 1 / C = 1 / ( 2 π f C )

= 1 / ( 2 × 3.14 × 30 × 65 × 10^-6 )

= 10⁶ / 12246

= 81.65 ohms.

Maximum Current I_max = 2 × I_rms

= 2 × ( V_rms / X_c )

= 1.414 × ( 14 / 81.65 )

= 1.414 × 0.171

= 0.242 A

= 242 mA

b) Voltage across the capacitor V = 4.30 V

Maximum Voltage V_max = 2 V_rms

= 1.414 × 14

= 19.796 V

Voltage across the capacitor V = V_max sin ( - 90)

4.3 = 19.796 sin ( - 90 )

sin ( - 90 ) = 4.3 / 19.796

sin ( - 90° ) = 0.217

   - 90 = sin^-1 ( 0.217 )

      = 90 + 12.53

= 102.53°

When the voltage is increasing = 102.53°

When the voltage is decreasing = 282.53°

Current in the circuit when voltage is increasing is

I = I_max sin

= 242 × 10^-3 × sin ( 102.53)

= 242 × 10^-3 × 0.976

= 236.19 mA

c) When the voltage is decreasing , current in the circuit is

I = I_max sin ( 282.53)

= 242 × 10^-3 × ( - 0.976 )

= - 236.19 mA