wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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?

Open in App
Solution

(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 =120πrad/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 =323A

Thus, the value of maximum current in the circuit is 323A.

(b)

In a capacitor circuit, voltage lags behind the current.

The equation at t=0 is,

ωtϕ=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 =06635 ϕ= 3356π 180 rad

Substitute the value of ϕ in equation (1).

t= 3356π 180×2π×60 t=155× 10 3 s =155ms

Thus, the time lag between maximum current and maximum voltage is 155ms.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Alternating Current Applied to a Capacitor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon