Question

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is $5\mathrm{A}$ and it is decreasing at a rate of ${10}^3{\mathrm{As}}^{-1}$ then ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}$ equals

Answer 1

Step 1: Given information

Current, $\mathrm{I}=5\mathrm{A}$

Decreasing rate = ${10}^3{\mathrm{As}}^{-1}$

Resistance, $\mathrm{R}=1\mathrm{\Omega }$

Voltage, $\mathrm{V}=15\mathrm{V}$

We have to find the number equivalent to ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}$.

Step 2: Calculation.

As we move from point A to point B,

We will construct the voltage drop equation.

${\mathrm{V}}_{\mathrm{A}}-\mathrm{iR}+\mathrm{V}+\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}={\mathrm{V}}_{\mathrm{B}}$

By substituting the values, we get

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}-5\times 1+15-5\times {10}^{-3}\times (-{10}^3)=\mathrm{VB} \\ {\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=15 \end{gathered}$$

Therefore, ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}$ equals $15$.