Question

An alternating emf of 220 V is applied to a circuit containing a resistor R having the resistance of 160Ω and a capacitor ‘C’ is series. The current is found to lead the supply voltage by an angle

`θ = tan^(-1) ("3/4")`

a) Calculate

1) The capacitive reactance

2) The impedance of the circuit

3) Current flowing in the circuit

b) If the frequency of the applied emf is 50 Hz, what is the value of the capacitance of the capacitor ‘C’?

Answer 1

a) E = 220 V, R = 160, C = Capacitor

Current leads voltage by `phi = tan^(-1) (3/4)`

Find `X_C, Z, I`

a) Given θ = `tan^(-1) (3/4)`  ...(1)

but θ = `tan^(-1) ((X_C)/R)` ...(2)

From 1 and 2

tan θ  = 3/4

but tan θ = `X_C/R`

`:. 3/4 = X_C/R` 

∵ R = 160

`:. 3/4= X_C/160`

`:. X_C = 3/4 xx 160`

`X_C = 120 Ω`

Impedance Z = `sqrt(R^2 + X_(C^2))`

`= [(160)^2 + (120)^2]^("1/2")`

Z = 200 Ω

`I_0 = V/Z = 220/2000 = 1.1A => I_0 = 1.1 A`

b) f = 50 Hz

`X_C = 120Ω`

`X_C = 1/(2pifC)`

`:. C = 1/((2pif)xx(X_C)) = 1/(2pi xx 50xx 120) = 2.65 xx10^(-5) F = 26.5 muF`

`C = 26.5 mu F`