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

Answer:

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 2

Given: A student is working on a circuit in which a resistance of 10 Ω is connected in series with an inductance across 200 V, 50 Hz in such a way that it takes 5 A current.

To find: The value of inductance of entire circuit and nature of inductive circuit with determination of power factor.​

Solution:

• For a circuit containing inductance and resistance in series,

        $I = \frac{V}{\sqrt{R^{2}+X_{L}^{2}} }$

• Here, I is the current, V is the voltage, R is the resistance and $X_{L}$ is the inductive reactance.

• $X_{L}$ is given by,

        $X_{L} = 2\pi fL$

• Here, f is the frequency, that is, 50 Hz and L is the inductance of the circuit to be found.

        $X_{L} = 2 * \pi * 50Hz * L$

              $= 100\pi L$

• So, the inductance is calculated as,

        $5 A = \frac{200 V}{\sqrt{ (10 ohm )^{2} + (100\pi L)^{2}} }$

        $L = 0.12 H$

• The power factor of the circuit can be calculated as,

        $cos \phi = \frac{R}{\sqrt{R^{2} + (X_{L})^{2}} }$

• Here, cos Ф is the power factor.

        $cos \phi = \frac{10}{\sqrt{10^{2} + (37.71)^{2}} }$

                $= 0.26$

Therefore, the value of inductance of the entire circuit is 0.12 H and the power factor is 0.26.