Question

In the adjoining circuit, initially the switch $\text{S}$ is open. The switch $\text{S}$ is closed at $t=0.$ The difference between the maximum and minimum currents that can flow in the circuit is-

• A. $2\text{A}$
• B. $3\text{A}$
• C. $1\text{A}$
• D. Nothing can be concluded.

Answer 1

The correct option is C $1\text{A}$

The battery will supply current in both the branches as shown in the figure below.

As $i_1$-passes through the inductor, it will vary with time and $i_2$ will be a constant current.

Hence, maximum current will be in the circuit when $i_1$ is maximum and vice-versa.


Net current through the circuit is, $i=i_1+i_2$

At any time current through the inductor (growth circuit) is,

$i=i_0[1-e^{(-\frac{t}{\tau })}]$

$\Rightarrow i_1=\frac{E}{R}[1-e^{(-\frac{t}{\tau })}]......\left(1\right)$

The time constant of the given circuit is,

$\tau =\frac{L}{R}=\frac{0.1}{10}=0.01\text{s}$

From (1) we get,

$i_1=\frac{10}{10}[1-e^{(-\frac{t}{0.01})}].......\left(2\right)$

From (2) we get, $i_1$ will be minimum at $t=0.$

$i_1$ is minimum at $t=0$.

$\Rightarrow (i_i{)}_{min}=1-e^0=0\text{A}$

$i_1$ is maximum at $t=\infty$.

$\Rightarrow (i_i{)}_{max}=1-e^{-\infty }=1-0=1\text{A}$

Now, current through the only resistor branch $i_2$ is,

$(i_2{)}_{max}=\frac{E}{R}=\frac{10}{10}=1\text{A}$

Therefore, the net maximum current flowing in the circuit is,

$i_{max}=(i_1{)}_{max}+i_2=1+1=2\text{A}$

Therefore, the net minimum current flowing in the circuit is,

$i_{min}=(i_1{)}_{min}+i_2=0+1=1\text{A}$

Therefore, the difference between $i_{max}$ and $i_{min}$ is,

$\therefore i_{max}-i_{min}=2-1=1\text{A}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.

Why this question ? Tip: $i$ will be maximum at $t=0$, when the inductor behaves like infinite resistance $(R=\infty )$ and minimum when it acts like a wire $(R=0)$.