Question

A 100 μF capacitor in series with a 40 W resistance is connected to a 110 V, 60 Hz supply. (a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum?

Answer 1

(a)

It is given that the capacitance of a capacitor, C=100μF, the resistance of the

Resistor, R=40Ω, supply voltage, V=110V and the frequency of supply,

ν=60Hz.

The formula of angular frequency is,

ω=2πν

Substitute the value of ν in the above equation.

ω=2π×60 =120π rad/s

The formula of maximum current in the circuit is,

I 0 = 2 V R 2 + 1 ω 2 C 2

Substitute the values.

I 0 = 2 ×110 ( 40 ) 2 + 1 ( 120π ) 2 × ( 10 −4 ) 2 I 0 =3⋅23 A

Thus, the value of maximum current in the circuit is 3⋅23 A.

(b)

In a capacitor circuit, voltage lags behind the current.

The equation at t=0 is,

ωt−ϕ=0 t= ϕ ω (1)

The formula of phase angle is,

tanϕ= 1 ωC R = 1 ωCR

Substitute the values.

tanϕ= 1 120π× 10 −4 ×40 =0⋅6635 ϕ= 33⋅56π 180  rad

Substitute the value of ϕ in equation (1).

t= 33⋅56π 180×2π×60 t=1⋅55× 10 −3 s =1⋅55 ms

Thus, the time lag between maximum current and maximum voltage is 1⋅55ms.