Question

A student is working on a circuit in which a resistance of 10Ω is connected in series with an inductance across 200V, 50Hz in such a way that it takes 5A current. Measure the value of inductance of entire circuit. Observe the nature of inductive circuit with determination of power factor.​

Answer 1

Given: R= 10Ω ,f= 50Hz, V= 200V ,I = 5A

To find: Inductance of the entire circuit(L)

Solution: since a resistance and an inductor is present in the circuit. Therefore, it is called a R-L circuit and for a R-L circuit,

I= V/Z

where I is current through the circuit, V is voltage across the circuit, Z is Impedance of the circuit.

By putting the given value of I and V

Z= 200/5= 40Ω

for a R-L circuit, Z= √( R^2+ (wL) ^2)

Z^2= R^2+ (wL) ^2

40^2= 10^2+ (wL) ^2

(wL) ^2= 1500

now f= 1/T and T= 2π/w

therefore f= w/2π

w= 2πf

w= 2π50

w= 100π

L= √(1500/10000π^2)

Therefore, the value of inductance of entire circuit will be √(1500/((100π)^2))

Power factor for a R-L circuit is cosϕ = R/Z

cosϕ = 10/40= 1/4

Therefore, power factor of the circuit is cosϕ= 1/4.