Question

In Figure, after switch S is closed at time$\mathit{t}\mathbf{=}\mathbf{0}$, the emf of the source is automatically adjusted to maintain a constant current i through S. (a) Find the current through the inductor as a function of time. (b) At what time is the current through the resistor equal to the current through the inductor?

Short answer

1. Current through the inductor is $i_2\left(t\right)=i\left(\begin{array}{c}1-e^{-\frac{Rt}{L}}\\ \end{array}\right)$
2. Time at which the current through the resistor is equal to the current through the inductor is$t=\frac{L}{R}In\left(2\right)$

Step-by-step solution

Given

Answer is missing

Understanding the concept

Here, we need to use the equations of Kirchhoff’s junction rule and loop rule. Using these equations, we can get the differential equation of the current through the resistor. The solution of this differential equation is the function of time, and from that, we can get the expression for the current through the inductor. The expression for a time at which the current through the resistor and inductor is equal can be found by equating the equations of current and solving for time.

Formula:

$$\begin{gathered} V_R=i_{1}R \\ {V}_L=L\times \frac{d{i}_2}{dt} \end{gathered}$$

(a) Find the current through the inductor as a function of time 

Applying junction rule to writ the expression for current in circuit:

$i=i_1+i_2$

Here I is the current from the constant current source.

$i_1$is current through the resistor R

$i_2$ is current through the inductor L

Taking derivative with respect to t of equation of junction rule, we get

$\frac{d{i}_1}{dt}=-\frac{d{i}_2}{dt}$

Now, $V_R$ is the voltage drop across resistor, and s is given by

$V_R=i_{1}R$

And$V_L$is voltage drop across inductor given as,

$V_L=L\times \frac{d{i}_2}{dt}$

Now applying voltage rule, we get

$$\begin{gathered} i_{1}R-L\times \frac{d{i}_2}{dt}=0 \\ {i}_{1}R+L\frac{d{i}_1}{dt}=0 \end{gathered}$$

This is differential equation, and its solution can be written as

role="math" localid="1661852425987" $i_1\left(t\right)=i_0{e}^{-\frac{Rt}{L}}$

where, $i_0$is the current in the circuit at time$=0$.

When switch is closed, the current through the inductor is zero. So,

$i_2=0$and $i=i_0$

So, the expression for current through the resistance as function of time is

$i_1\left(t\right)=i{e}^{-\frac{Rt}{L}}$

The current through the inductor starts to rise as time passes, so $i_2$is as follows:

role="math" localid="1661852573179" $$\begin{gathered} i_2\left(t\right)=i-i_1\left(t\right) \\ {i}_2\left(t\right)=i-i{e}^{-\frac{Rt}{L}} \\ {i}_2\left(t\right)=i\left(1-e^{-\frac{Rt}{L}}\right) \end{gathered}$$

(b) Calculate time at which the current through the resistor is equal to the current through the inductor.

The current through resistor and inductor is equal. So,

$$\begin{gathered} i_1\left(t\right)=i_2\left(t\right) \\ i\left(1-e^{-\frac{Rt}{L}}\right)=i{e}^{-\frac{Rt}{L}} \\ \left(1-e^{-\frac{Rt}{L}}\right)=e^{-\frac{Rt}{L}} \\ {e}^{-\frac{Rt}{L}}=\frac{1}{2} \\ -\frac{Rt}{L}=-In\left(2\right) \\ t=\frac{L}{R}In\left(2\right) \end{gathered}$$