Question

An inductor with an inductance of $3.00H$ and a resistance of $7.00\Omega$ is connected to the terminals of a battery with an emf of $12.0V$ and negligible internal resistance. What is the power input to the inductor from the battery as a function of time if the circuit is completed at $t=0$?

Answer 1

Fig: R-L circuit

Given,

$L=3.00H$

$R=7.00\Omega$

$V_s=12.0V$

The current in the LR circuit as a function of time,

$I=I_0(1-e^{-t/\tau })$

where,$\tau =$ Time constant

$=\frac{L}{R}$

The power input to the inductor from the battery as a function of time,

$P=VI$

$P=LI\frac{dI}{dt}$

$P=LI\frac{d}{dt}{I}_0(1-e^{-t/\tau })$

$P=I_0^{2}R{e}^{-t/\tau }(1-e^{-t/\tau })$

The circuit is completed at $t=0$

$P=\frac{V_s^2}{R}$

$P=\frac{12\times 12}{7}W$

$P=20.5714W$