Question

At time t = 0, a 12.0 V potential difference is suddenly applied to the leads of a coil of inductance 23.0 mHand a certain resistance R. At time t = 0.150 ms, the current through the inductor is changing at the rate of 280 A/s. Evaluate R.

Short answer

The value for R will be$95\Omega$.

Step-by-step solution

Given

• $L=0.023\text{H}$.
• $\epsilon =12\text{V}$.
• $t=0.00015$.
• $\frac{di}{dt}=280\text{A/s}$.

Understanding the concept

We have the equation for the rise in the current. We can take the derivative for the current to evaluate the value for R.

Formula:

$i=\frac{\epsilon }{R}\left(1-e^{-t/{\tau }_L}\right)$

${\tau }_L=\frac{L}{R}$

Step 3: Evaluate the value for R

We have,

$i=\frac{\epsilon }{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)$

We take derivative for both sides with respect to t, we get,

$\frac{di}{dt}=\frac{d\left(\frac{\epsilon }{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)\right)}{dt}$

We have,

$\begin{array}{rcl}{\tau }_L& =& \frac{L}{R}\\ \frac{di}{dt}& =& \frac{d\left(\frac{\epsilon }{R}\left(1-e^{-\frac{t}{\left(\frac{L}{R}\right)}}\right)\right)}{dt}\\ \frac{di}{dt}& =& \frac{\epsilon R}{RL}\times e^{-\frac{t}{\left(\frac{L}{R}\right)}}\\ \frac{di}{dt}& =& \frac{\epsilon }{L}\times e^{-\frac{t}{\left(\frac{L}{R}\right)}}\\ \frac{L}{\epsilon }\times \frac{di}{dt}& =& e^{-\frac{t}{\left(\frac{L}{R}\right)}}\\ e^{\frac{-Rt}{L}}& =& \frac{0.023\times 280}{12}\\ e^{\frac{-Rt}{L}}& =& 0.537\\ & & \end{array}$

Taking natural log at both sides, we get,

$\begin{array}{rcl}-\frac{Rt}{L}& =& \log 0.536\\ \frac{Rt}{L}& =& 0.62\\ R& =& \frac{0.62\times L}{t}\\ & =& \frac{0.62\times 0.023}{0.00015}\\ & =& 95.0\Omega \\ & & \end{array}$

Therefore, the value of resistance is $95.0\Omega$