Question

The current in an RL circuit drops from1.0 A to 10 mA in the first second following removal of the battery from the circuit. If L is10 H, find the resistance R in the circuit.

Short answer

$R=46\Omega$

Step-by-step solution

Given data 

• $i_0=1\text{A}$
• $i=10\times {10}^{-3}\text{A}$
  

Understanding the concept

When the source of constant emf is removed, the current decays from a valueaccording to,

$\mathit{i}\mathbf{=}{\mathit{i}}_{\mathbf{0}}{\mathit{e}}^{\frac{\mathbf{-}\mathbf{t}}{{\mathbf{\tau }}_{\mathbf{L}}}}$

Formulae:

The current decay in the RL circuit is given by,

$i=i_0{e}^{\frac{-t}{{\tau }_L}}$

The inductive time constant is,

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

Calculate the resistance R in the circuit 

The current decay inRLcircuit is given by,

$\begin{array}{rcl}i& =& i_0{e}^{\frac{-t}{{\tau }_L}}\\ \ln \left(\frac{i}{i_0}\right)& =& \frac{-t}{{\tau }_L}\\ {\tau }_L& =& \frac{-t}{\ln \left(\frac{i}{i_0}\right)}\\ & & \end{array}$

Hence,

$\frac{L}{R}=\frac{-t}{\ln \left(\frac{i}{i_0}\right)}$

$\frac{10}{R}=\frac{-1}{\text{ln}\left(\frac{10\times {10}^{-3}}{1.0}\right)}$

By solving for R,

$R=46\Omega$

Therefore the resistance in the circuit is 46 ohm.