In the circuit shown, The switch S1 is closed at time t = 0 and the switch S2 is kept open. At some

Q: Why doesn't the current reach its maximum value instantly when the switch is closed?

The inductor creates a back EMF (electromotive force) that opposes any change in current. This opposition causes the current to rise gradually rather than instantaneously.

Q: What happens to the energy stored in the inductor after $t_0$?

During the discharging phase, the stored magnetic energy in the inductor is converted into electrical energy, which is then dissipated as heat in the resistor.

Q: Why does the current decay exponentially instead of linearly?

The rate of decay of current is proportional to the current itself ($\frac{di}{dt} \propto -i$). Mathematically, a quantity whose rate of change is proportional to its current value always follows an exponential function.

Q: What is the time constant of this circuit, and what does it mean?

The time constant is $\tau = \frac{L}{R}$. It is a measure of how fast the current grows or decays. A larger time constant means the current changes more slowly.

Q: Would the shape of the graph change if the resistance was higher?

Yes, a higher resistance would decrease the time constant ($\tau = \frac{L}{R}$). This means both the charging and discharging phases would happen faster, resulting in steeper curves on the graph.

Question

In the circuit shown,

The switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time ($t_0$), the switch $S_1$ is opened and $S_2$ is closed. The behaviour of the current $I$ as a function of time '$t$' is given by

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's carefully analyze the circuit provided in the problem. We have an inductor $L$ and a resistor $R$ connected in series in the top branch. There are two switches, $S_1$ and $S_2$, which control the flow of current through different parts of the circuit.

Initially, at time $t = 0$, switch $S_1$ is closed and $S_2$ is kept open. This configuration connects the battery (with electromotive force $\varepsilon$) directly to the $L-R$ branch, forming a complete loop. This is the classic setup for the charging of an inductor.

### The Charging Phase ($t \le t_0$)

When $S_1$ is closed, the battery attempts to drive a current through the circuit. However, the inductor opposes any sudden change in current due to its self-inductance. As a result, the current doesn't jump to its maximum value instantly. Instead, it grows exponentially over time.

The equation governing the current during this charging phase is given by:

$$I = \frac{\varepsilon}{R} \left[ 1 - \exp\left(-\frac{R}{L}t\right) \right]$$

Here, $\frac{\varepsilon}{R}$ represents the maximum steady-state current that would eventually flow if the circuit were left undisturbed for a long time. The term $\frac{L}{R}$ is the time constant of the circuit, which determines how quickly the current rises. This exponential growth continues until time $t_0$.

### The Discharging Phase ($t > t_0$)

At time $t_0$, a sudden change occurs: switch $S_1$ is opened, and switch $S_2$ is closed simultaneously. Opening $S_1$ disconnects the battery from the circuit. Closing $S_2$ creates a new closed loop consisting only of the inductor $L$ and the resistor $R$.

Now, the inductor, which has stored magnetic energy during the charging phase, acts as a source. It drives a current through the resistor to maintain the flow. Without the battery to sustain it, the current begins to decay. The energy stored in the magnetic field of the inductor is gradually dissipated as heat in the resistor.

The current during this discharging phase decays exponentially according to the equation:

$$I = I(t_0) \exp\left[-\frac{R}{L}(t - t_0)\right]$$

where $I(t_0)$ is the current that had been established in the circuit exactly at time $t_0$, and $(t - t_0)$ is the time elapsed since the switches were flipped.

### Conclusion

Combining our observations from both phases, the complete behavior of the current $I$ as a function of time $t$ consists of an **exponential growth** from $t = 0$ to $t = t_0$, followed immediately by an **exponential decay** for $t > t_0$.

When we examine the given options, the graph that perfectly illustrates this two-part exponential behavior is option (b).

Select Answer:

Visualized Solution

Initial Setup

Charging of Inductor

Charging Equation

Switching State

Discharging of Inductor

Discharging Equation

Final Graph

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Charging Phase ( $t \leq t_{0}$ )

The Discharging Phase ( $t > t_{0}$ )

Conclusion

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