Question

An inductor of `5/pi` H, a capacitor of `50/pi` µF and a resistor of 400 Ω are connected in series across an ac voltage v = 140 sin(100 πt) V. Calculate:

1. impedance of the circuit, and
2. rms value of current that flows in the circuit. (Take `sqrt 2` = 1.4).

Answer 1

Given: L = `5/pi` H

C = `50/pi` µF = `50/pi xx 10^-6` F

R = 400 Ω

V = 140 sin(100 πt)

From the voltage equation:

V₀ = 140 V and

ω = 100 π rad/s

Inductive Reactance (X_L) = ωL

= `100 pi xx 5/pi`

= 500 Ω

Capacitive Reactance (X_C) = `1/(omega C)`

= `1/(100 pi xx 50/pi xx 10^-6)`

= `1/(5000 xx 10^-6)`

= `10^6/5000`

= 200 Ω

I. Impedance (Z) = `sqrt(400^2 + (500 - 200)^2)`

= `sqrt(160000 + 90000)`

= `sqrt 250000`

= 500 Ω

II. RMS Current (I_rms):

Peak current (I₀) = `V_0/Z`

= `140/500`

= 0.28 A

I_rms = `I_0/sqrt 2`

= `0.28/1.4`

= 0.2 A