Question

An inductor with an inductance of $2.50H$ and a resistance of $8.00\Omega$ is connected to the terminals of battery with an emf of $6.00V$ and negligible internal resistance. Find the rate of increase of current at the instant when the current is $0.500A$.

Answer 1

$V_L=E-iR$
$⟹L\frac{di}{dt}=E-iR$
$⟹\frac{di}{dt}=\frac{E-iR}{L}$
$⟹\frac{di}{dt}=\frac{6-0.5\times 8}{2.5}=0.8A/s$