Question

The power factor in the circuit is 1/root2 . The capacitance of the circuit is equal to

(a) 400 mF

(b) 300 mF

(c) 500 mF

(d) 200 mF

Answer 1

Answer:

Explanation:

=> Here, it is given that,

power factor in the circuit, = 1/√2

Resistance, R = 10Ω

inductance, L = 0.1 H

Supply voltage, V = 2sin(100t)

peak voltage, V₀ = 2

ω = 100 rad/s

capacitance = ?

=> power factor = cos∅ = R/Z

1/√2 = 10/Z

=> By Taking the square on both the sides,

1/2 = 100/Z²

Z² = 200

=> Now, we can use below formula to calculate the capacitance of the circuit.

R² + (Xc - XL)² = Z²

10² + (1 / cω - Lω)² = 200

(1/cω - Lω)² = 100

1/cω - Lω = 10

1/cω = Lω + 10

1/cω = 0.1 * 100 + 10

1/cω = 20

c = 1/20*ω

c = 1/20*100

c = 1/2000 F

c = 10⁶ / 2000 (μ)F    [∵ 1 F = 10⁶ (μ)F ]

= 500 (μ) F

Thus, the capacitance of the circuit is equal to 500 (μ) F.