Question

The time constant of a circuit is 10 sec, When a resistance of $100\Omega$ is connected in series in a previous circuit then time constant becomes 2 second,then the self inductance of the circuit is;-

• A. $250H$
• B. $50H$
• C. $150H$
• D. $25H$

Answer 1

The correct option is A $250H$

In LR circuit,

The time constant $\tau =\frac{L}{R}$

$10=\frac{L}{R}$

$L=10R$. . . . . . .(1)

When Resistance $100\Omega$ is connect in series, than the time constant is

${\tau }^{\prime }=\frac{L}{R+100}=2s$

$L=2R+200$. . . . . . .(2)

Equating equation (1 ) and (2), we get

$2R+200=10R$

$8R=200$

$R=25\Omega$

From equation (1),

$L=10R=10\times 25$

$L=250H$