Question

The variation of inductive reactance (X_L) of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in fig.

1. Calculate the self-inductance of the inductor.
2. When this inductor is used in series with a capacitor of unknown value and a resistor of 10 Ω at 300 s^–1, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.

Answer 1

i. We know that X_L = ωL and ω = 2π f

Where,

f = frequency in Hz

So,  L = `X_L/(2 pi f)`

= `(20)/(2 pi(100))`

= `(40)/(2pi (200))`

= `(60)/(2pi(300))`

= 31.84 × 10⁻³

≈ 32 mH

ii. We know that power dissipation is maximum when,

X_L = X_C

⇒ ωL = `1/(omegaC)`

⇒ `C = 1/(omega^2L)`

⇒ C = `1/(4 pi^2 f^2 L)`

= `1/(4 xx 3.14 xx 3.14 xx 300 xx 300 xx 32 xx 10^-3)` 

= 8.8 μF

Answer 2

i. Given: From graph f = 100 Hz

X_L = 20 Ω

Inductive reactance (X_L)  = 2π f L

So,

L = `X_L/(2 pi f)` 

= `(20)/(2 pi xx 100)`

 = 0.032 H

= 32 mH

ii. Given: f = 300 s⁻¹

L = 0.032H

We know that power dissipation is maximum when, 

2π f L = `1/(2 pi f C)`

2π × 300 × 0.032 = `1/(2 pi xx 300 xx C)`

∴ C = 8.8 × 10⁻⁶ F

= 8.8 μF