Question

In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45° is `1/x xx 10^-3` F. Then the value of x is ______.

विकल्प

• 1

• 2

• 3

• 4

Answer 1

In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45° is `1/x xx 10^-3` F. Then the value of x is 3.

Explanation:

Inductor (L) = 30 mH = 30 × 10^-3 H

Resistor (R) = 1 Ω

Angular frequency ω = 300 rad/s

Angular phase Φ = 45°

Use this expression,

`tanPhi = (X_C - X_L)/R = 1`

X_C - X_L = R

`1/(omegaC) - omegaL = R`

`1/(omegaC) = R + omegaL`

⇒ `1/(omegaC) = 1 + 300(30 xx 10^-3) = 10 Ω`

So, `C = 1/(10omega)`

= `1/(10 xx 300) = 1/3 xx 10^-3`F

x = 3