Question

A current of $4$A flows in a coil when connected to a $12$V dc source. If the same coil is connected to a $12$V, $50rad/s$ a.c. source, a current of $2.4$A flows in the circuit. Determine the inductance of the coil.

Answer 1

In case of DC 12 V let r is internal resistance of coil and l is the inductance of coil of steady state inductor behave short circuit.

$Idc=\frac{12}{r}=4$

$\Rightarrow r=3\Omega$

In AC current flowing = 2.4 A

Applying KVL

$12=3L+j{X}_{L}12=3\times 2.4+j(2.4\times \omega L)12=7.2+j2.4\times 5DL=7.2+j12L12=7.2+j12L4=2.4+j4L4=\sqrt{\left(2.4{)}^2\right(4L{)}^2}=\sqrt{5.76+16L^2}16=5.76+16L^2{L}^2=\frac{10.24}{16}=0.64L=0.8H$