The time constant of an LR circuit is 10 , s. | vedclass

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(D) The time constant $	au$ of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$\frac{L}{R} = 10$,which implies $L = 10R$  --- $(1)$.When

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$25$

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(D) The time constant $	au$ of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$\frac{L}{R} = 10$,which implies $L = 10R$  --- $(1)$.When a resistance of $10 \, \Omega$ is added in series,the new resistance becomes $(R + 10) \, \Omega$. The new time constant is $\frac{L}{R + 10} = 2$,which implies $L = 2(R + 10) = 2R + 20$ --- $(2)$.Equating $(1)$ and $(2)$:$10R = 2R + 20$$8R = 20$$R = 2.5 \, \Omega$.Substituting $R$ into equation $(1)$:$L = 10 	imes 2.5 = 25 \, H$.

The time constant of an $LR$ circuit is $10 \, s$. When a resistance of $10 \, \Omega$ is connected in series with the existing circuit,the time constant becomes $2 \, s$. The self-inductance of the circuit is ..... $H$.

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