The network shown in figure is part of a complete circuit. If at a certain instant, the current i is 5A and is decreasing at a rate of ${10}^3\mathrm{A}/\mathrm{s}$, then ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=5\mathrm{k}$ volts, find the value of k.

Given, $\frac{\mathrm{di}}{\mathrm{dt}}={10}^3\mathrm{A}/\mathrm{s}$

$\therefore$ Induced emf across inductance, $\left|\mathrm{e}\right|=\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}$

So, $\left|\mathrm{e}\right|=(5\times {10}^{-3})\left({10}^3\right)\mathrm{V}=5\mathrm{V}$

Since, the current is decreasing, the polarity of this emf would be so as to increase the existing current. The circuit can be redrawn as

Now, ${\mathrm{V}}_{\mathrm{A}}-5+15+5={\mathrm{V}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-15\mathrm{V}\text{ or }{\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=15\mathrm{V}=5\times 3\mathrm{V}$