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

${\displaystyle \theta ={\tan }^{-1}\left(\text{3/4}\right)}$

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 ${\displaystyle \varphi ={\tan }^{-1}\left(\frac{3}{4}\right)}$

Find ${\displaystyle X_C,Z,I}$

a) Given θ = ${\displaystyle {\tan }^{-1}\left(\frac{3}{4}\right)}$  ...(1)

but θ = ${\displaystyle {\tan }^{-1}\left(\frac{X_C}{R}\right)}$ ...(2)

From 1 and 2

tan θ  = 3/4

but tan θ = ${\displaystyle \frac{X_C}{R}}$

${\displaystyle \therefore \frac{3}{4}=\frac{X_C}{R}}$ 

∵ R = 160

${\displaystyle \therefore \frac{3}{4}=\frac{X_C}{160}}$

${\displaystyle \therefore X_C=\frac{3}{4}\times 160}$

${\displaystyle X_C=120 \Omega }$

Impedance Z = ${\displaystyle \sqrt{R^2+X_{C^2}}}$

${\displaystyle ={[{\left(160\right)}^2+{\left(120\right)}^2]}^{\text{1/2}}}$

Z = 200 Ω

${\displaystyle I_0=\frac{V}{Z}=\frac{220}{2000}=1.1A\Rightarrow I_0=1.1A}$

b) f = 50 Hz

${\displaystyle X_C=120\Omega }$

${\displaystyle X_C=\frac{1}{2\pi fC}}$

${\displaystyle \therefore C=\frac{1}{\left(2\pi f\right)\times \left(X_C\right)}=\frac{1}{2\pi \times 50\times 120}=2.65\times {10}^{-5}F=26.5\mu F}$

${\displaystyle C=26.5\mu F}$