Question

Find the self-inductance $L$ (in $mH$ ) of the coil whose resistance $R=2 \,ohm$, a current $i=1.5 \,amp$, which is increasing at a rate of $10^{3} \, amp / s$.

Answer 1

The current is increasing and flowing left to right in the inductor then left end of inductor will have high potential and right end of inductor will be at low potential. Applying Kirchhoff's loop law
$\varepsilon-V_{L}-i R=0 \Rightarrow \varepsilon-L \frac{d i}{d t}-i R=0$


$6.0-L \times 10^{3}-1.5 \times 2=0$
$\Rightarrow L=\frac{6.0-3.0}{10^{3}}=3.0 \times 10^{-3} H$,
Hence $L=3 \,mH$