The applied potential difference in the circuit shown is V = 10 sin 100 t where V is in volt and t is in second. If the power factor of the circuit is 1/√2, then the value of capacitance C of the circ

Question

The applied potential difference in the circuit shown is V = 10 sin 100 t where V is in volt and t is in second. If the power factor of the circuit is 1/√2, then the value of capacitance C of the circuit, nearly is

Accepted Answer

Answer: 333 μF — XL = ωL = 100×0.1 = 10Ω For power factor 1/√2: XC = 30Ω C = 1/(ωXC) = 1/(100×30) = 333 μF

Practice Physics Exams