For the given input voltage waveform V_in(t), the output voltage waveform V_out(t), across the capac

Q: Why doesn't the capacitor charge fully to 5V during the first pulse?

The capacitor needs about 5 time constants ($5\tau$) to fully charge. Here, $\tau = 10\,\mu\text{s}$, but the pulse only lasts for $5\,\mu\text{s}$ (which is $0.5\tau$). It simply doesn't have enough time to reach 5V before the input drops back to zero.

Q: How do I quickly estimate the value of $e^{-0.5}$ without a calculator?

It's highly recommended to memorize a few key values of $e^{-x}$ for JEE. $e^{-1} \approx 0.37$, and $e^{-0.5} = 1/\sqrt{e} \approx 0.606$. Knowing this makes these RC circuit problems much faster to solve.

Q: Why does the second charging phase use a different formula?

The standard formula $V_m(1 - e^{-t/\tau})$ assumes the capacitor starts completely uncharged ($0\text{ V}$). In the second charging phase, the capacitor already holds about $1.2\text{ V}$. You must use the general formula $V(t) = V_{final} + (V_{initial} - V_{final})e^{-t/\tau}$ to account for this initial charge.

Q: What would happen if the input stayed at 5V forever?

If the input remained at 5V, the exponential term $e^{-t/\tau}$ would eventually approach zero as time $t$ gets very large. The voltage across the capacitor would asymptotically approach the maximum input voltage of 5V.

Q: How do we know the graph is an exponential curve and not a straight line like in option (b)?

The rate at which a capacitor charges depends on the voltage difference between the source and the capacitor. As the capacitor charges, this difference decreases, so the charging current decreases, leading to a curve that flattens out (an exponential curve). A straight line would imply a constant charging current, which only happens if you charge a capacitor with a constant current source, not a constant voltage source.

Question

For the given input voltage waveform $V_{in}(t)$, the output voltage waveform $V_{out}(t)$, across the capacitor is correctly depicted by

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a dynamic electrical system. We have a simple yet fascinating RC circuit consisting of a $1	ext{ k}\Omega$ resistor and a $10	ext{ nF}$ capacitor connected in series. The driving force behind this circuit is a square wave voltage source, $V_{in}(t)$, which acts like a switch, alternating between $5	ext{ V}$ and $0	ext{ V}$ every $5\,\mu	ext{s}$. Our mission is to trace the exact path of the output voltage, $V_{out}(t)$, across the capacitor as it responds to these sudden changes.

### The Master Equation: Time Constant

Before we dive into the charging and discharging phases, we must determine the circuit's "heartbeat"—its time constant, $	au$. The time constant dictates how rapidly the capacitor can respond to changes in the input voltage.

$$	au = RC$$

Let's substitute the given values:

$$	au = (1 	imes 10^3\,\Omega) 	imes (10 	imes 10^{-9}	ext{ F}) = 10 	imes 10^{-6}	ext{ s} = 10\,\mu	ext{s}$$

This tells us that it takes $10\,\mu	ext{s}$ for the capacitor to charge to about $63\%$ of its maximum capacity. Since our input voltage changes every $5\,\mu	ext{s}$ (which is only half of $	au$), the capacitor will never have enough time to fully charge or fully discharge in a single cycle.

### Phase 1: The Initial Charge

For the first $5\,\mu	ext{s}$ ($0 \le t \le 5\,\mu	ext{s}$), the input voltage is a steady $5	ext{ V}$. The capacitor begins to charge from zero. The voltage across it follows the classic exponential charging curve:

$$V_{out}(t) = V_m(1 - e^{-t/	au})$$

At the end of this phase ($t = 5\,\mu	ext{s}$), the voltage reaches:

$$V_{out}(5\,\mu	ext{s}) = 5(1 - e^{-5/10}) = 5(1 - e^{-0.5})$$

Knowing that $e^{-0.5} \approx 0.606$, we can calculate:

$$V_{out}(5\,\mu	ext{s}) \approx 5(1 - 0.606) = 5(0.394) = 1.97	ext{ V}$$

So, the voltage climbs to approximately $1.97	ext{ V}$ (or roughly $2	ext{ V}$).

### Phase 2: The Discharge

Suddenly, at $t = 5\,\mu	ext{s}$, the input voltage drops to $0	ext{ V}$. The capacitor, now holding $1.97	ext{ V}$, begins to discharge through the resistor. The discharge equation is:

$$V_{out}(t) = V_0 e^{-t'/	au}$$

Here, $V_0 = 1.97	ext{ V}$, and $t'$ is the time elapsed since the discharge began. At $t = 10\,\mu	ext{s}$, the elapsed time $t'$ is $5\,\mu	ext{s}$:

$$V_{out}(10\,\mu	ext{s}) = 1.97 e^{-5/10} = 1.97 e^{-0.5}$$

$$V_{out}(10\,\mu	ext{s}) \approx 1.97 	imes 0.606 \approx 1.19	ext{ V}$$

The voltage drops, but not all the way to zero. It settles at approximately $1.2	ext{ V}$.

### Phase 3: Re-charging from a Baseline

At $t = 10\,\mu	ext{s}$, the input voltage jumps back to $5	ext{ V}$. The capacitor starts charging again, but this time, it already has a "head start" of $1.19	ext{ V}$. The general equation for charging from an initial voltage $V_0$ is:

$$V_{out}(t) = V_m - (V_m - V_0) e^{-t'/	au}$$

At $t = 15\,\mu	ext{s}$, the elapsed time $t'$ is again $5\,\mu	ext{s}$:

$$V_{out}(15\,\mu	ext{s}) = 5 - (5 - 1.19) e^{-5/10}$$

$$V_{out}(15\,\mu	ext{s}) = 5 - 3.81 e^{-0.5} \approx 5 - 3.81(0.606)$$

$$V_{out}(15\,\mu	ext{s}) \approx 5 - 2.31 = 2.69	ext{ V}$$

### Final Conclusion

By tracking the voltage, we see a distinct pattern: it rises to $\approx 2	ext{ V}$, decays to $\approx 1.2	ext{ V}$, and then rises higher to $\approx 2.7	ext{ V}$. Comparing this mathematical reality to the given options, **Graph (a)** is the only one that correctly depicts this specific exponential growth and decay sequence.

Animated Solution for Physics - Current Electricity: For the given input voltage waveform $V_{in} (t)$ , the output voltage waveform $V_{o u t} (t)$ , across the capacitor is correctly depicted by

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Visualized Solution

Visualizing the Setup

Time Constant $τ$

Calculating $τ$

Phase 1: Charging ( $0$ to $5 μ s$ )

Voltage at $t = 5 μ s$

Phase 2: Discharging ( $5$ to $10 μ s$ )

Voltage at $t = 10 μ s$

Phase 3: Re-charging ( $10$ to $15 μ s$ )

Voltage at $t = 15 μ s$

Conclusion

The Way Forward

The Sigma Insight: RC Circuit

Analyzing the Setup

The Master Equation

Phase 1

Phase 2

Phase 3

Final Conclusion

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