Chapter 30: Q78P (page 899)

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

Expert verified

The value for R will be $95 Ω$ .

Step by step solution

Given

$L = 0.023 H$ .
$ε = 12 V$ .
$t = 0.00015$ .
$\frac{d i}{d t} = 280 A/s$ .

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:

$i = \frac{ε}{R} (1 - e^{- t / τ_{L}})$

$τ_{L} = \frac{L}{R}$

Step 3: Evaluate the value for R

We have,

$i = \frac{ε}{R} (1 - e^{- \frac{t}{τ_{L}}})$

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

$\frac{d i}{d t} = \frac{d (\frac{ε}{R} (1 - e^{- \frac{t}{τ_{L}}}))}{d t}$

We have,

$\begin{array}{rcl} τ_{L} & = & \frac{L}{R} \\ \frac{d i}{d t} & = & \frac{d (\frac{ε}{R} (1 - e^{- \frac{t}{(\frac{L}{R})}}))}{d t} \\ \frac{d i}{d t} & = & \frac{ε R}{R L} \times e^{- \frac{t}{(\frac{L}{R})}} \\ \frac{d i}{d t} & = & \frac{ε}{L} \times e^{- \frac{t}{(\frac{L}{R})}} \\ \frac{L}{ε} \times \frac{d i}{d t} & = & e^{- \frac{t}{(\frac{L}{R})}} \\ e^{\frac{- R t}{L}} & = & \frac{0.023 \times 280}{12} \\ e^{\frac{- R t}{L}} & = & 0.537 \end{array}$

Taking natural log at both sides, we get,

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