Question

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?

Short answer

0.520 ms is the time at which the emf across the inductor is equal to the potential difference across the resistor.

Step-by-step solution

Given

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

Understanding the concept

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:

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

Calculate at what time the emf across the inductor is equal to the potential difference across the resistor

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

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