Switch S in Fig. 30-63 is closed at time t = 0, initiating the buildup of current in the L = 15.0 mHinductor and the R = 20.0$\mathbf{\Omega }$resistor. At what time is the emf across the inductor equal to the potential difference across the resistor?

$$\begin{gathered} \mathrm{L}=15.0\mathrm{mH}=15.0\times {10}^{-3}\mathrm{H} \\ \mathrm{R}=20.0\mathrm{\Omega } \end{gathered}$$

We can find the equation for the emf across the inductor and potential difference across the resistor. We use the given condition and equate the m to find the required answer.

Formula:

$$\begin{gathered} V_R=iR \\ i=i_0\left[1-e^{-\left(R/L\right)t}\right] \\ {V}_L=L\frac{di}{dt} \end{gathered}$$

We have current in the circuit at any time when the switch is closed,

$i=i\left[1-e^{-\left(\frac{R}{L}\right)t}\right]$

Differentiating above equation with respect to t, we get.

$$\begin{gathered} \frac{di}{dt}=\frac{d\left(i_0\left[1-e^{-\left(\frac{R}{L}\right)t}\right]\right)}{dt} \\ \frac{di}{dt}=i_0\left(\frac{R}{L}\right)t\left[e^{-\left(\frac{R}{L}\right)t}\right] \end{gathered}$$

Voltage across the resistor can be given as,

$V_R=iR$

We substitute the value for

$$\begin{gathered} V_R=R\times \left(i_0\left[1-e^{-\left(\frac{R}{L}\right)t}\right]\right) \\ V=R{i}_0\left[1-e^{-\left(\frac{R}{L}\right)t}\right] \end{gathered}$$

We have,

$V_L=L\frac{di}{dt}$

We substitute the value for$\frac{di}{dt}$

$$\begin{gathered} V_L=L\times \left(i_0\left[1-e^{-\left(\frac{R}{L}\right)t}\right]\right) \\ {V}_L=R{i}_0\left[e^{-\left(\frac{R}{L}\right)t}\right] \end{gathered}$$

According to the given data,

$V_L=V_R$

We substitute the values for $V_L$and $V_R$

$$\begin{gathered} R{i}_0\left[e^{-\left(\frac{R}{L}\right)t}\right]=R{i}_0\left[1-e^{-\left(\frac{R}{L}\right)t}\right] \\ \left[e^{-\left(\frac{R}{L}\right)t}\right]=\left[1-e^{-\left(\frac{R}{L}\right)t}\right] \\ 2e^{-\left(\frac{R}{L}\right)t}=1 \\ {e}^{-\left(\frac{R}{L}\right)t}=\frac{1}{2} \end{gathered}$$

We take natural log at both sides

$$\begin{gathered} -\left(\frac{R}{L}\right)t=In\left(0.5\right) \\ \left(\frac{R}{L}\right)t=In0.693 \\ t=\frac{L\times 0.693}{R} \\ t=\frac{15.0\times {10}^{-3}\times 0.693}{R} \\ t=0.51975\times {10}^{-3}s\approx 0.520ms \end{gathered}$$