Question

The current (in Ampere) in an inductor is given by $I=5+16t$, where $t$ is in seconds. The self-induced emf in it is $10mV$. Find the energy stored in the inductor and the power supplied to it at $t=1s$.

Answer 1

$\frac{dI}{dt}=16A/s$

$\therefore L=\left|\frac{e}{\frac{dI}{dt}}\right|=\frac{10\times {10}^{-3}}{16}=0.625\times {10}^{-3}H$

$=0.625mH$

At $t=1s,I=21A$

$U=\frac{1}{2}L{I}^2=\frac{1}{2}(0.625\times {10}^{-3}){\left(21\right)}^2=0.137J$

$P=Ei=(10\times {10}^{-3})\left(21\right)=0.21J/s$