Question

In Fig. $\mathbf{30}\mathbf{.}\mathbf{11}$switch$\mathit{S}\mathbf{1}$is closed while switch $\mathit{S}\mathbf{2}$is kept open. The inductance is $\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{38}\mathit{H}$, the resistance is $\mathit{R}\mathbf{=}\mathbf{48}\mathit{\Omega }$, and the emf of the battery is $\mathbf{18}\mathit{V}$. At time $\mathit{t}$after $\mathit{s}\mathbf{1}$is closed, the current in the circuit is increasing at a rate of $\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{2}\mathit{A}\mathbf{/}\mathit{s}$. At this instant what is ${\mathit{v}}_{\mathbf{a}\mathbf{b}}$, the voltage across the resistor?

Short answer

Voltage across the resistance is $15.3V$.

Step-by-step solution

Calculate the voltage across the resistance

Consider the following circuit where $L=0.38H,R=48\Omega$and $\epsilon =18V$.

It is given that the switch $s_1$is closed while the switch $S_2$is kept open. At a certain instant the current is increasing at a rate of $\frac{di}{dt}=7.2A/s$, we need to find the voltage across the resistor at this instant. To find the potential, we apply the Kirchhoff’s loop rule to the upper loop, so we get,

$\epsilon -v_R-v_L=0$

But,$v_L=L\frac{di}{dt}$

So, localid="1664178496830" $\epsilon -v_R-L\frac{di}{dt}=0$

Solve for and substitute to get,

So, voltage across the resistance is $15.3V$.