During charging a capacitor variation of potential V of the capacitor with time t is shown as

Options

A.
B.
C.
D.

Explanation

To solve the problem regarding the variation of potential \( V \) of a capacitor with time \( t \) during charging, we can follow these steps:

### Step 1: Understand the Charging of a Capacitor
When a capacitor is charged through a resistor, the voltage across the capacitor increases over time according to an exponential function. The relationship can be described by the formula:

\[ V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) \]

where:
- \( V(t) \) is the potential across the capacitor at time \( t \),
- \( V_0 \) is the maximum potential (or supply voltage),
- \( R \) is the resistance in the circuit,
- \( C \) is the capacitance of the capacitor,
- \( e \) is the base of the natural logarithm.

### Step 2: Identify the Nature of the Graph
From the formula, we can see that as time \( t \) increases, the term \( e^{-\frac{t}{RC}} \) decreases, causing \( V(t) \) to approach \( V_0 \). This means that the graph of \( V \) versus \( t \) will start at \( 0 \) volts when \( t = 0 \) and will asymptotically approach \( V_0 \).

### Step 3: Determine the Shape of the Curve
The graph of \( V(t) \) will be an exponential curve. Initially, it will rise steeply and then gradually level off as it approaches \( V_0 \). This characteristic shape is typical of exponential growth.

### Step 4: Choose the Correct Option
Given the options (A, B, C, D), we need to identify which option represents an exponential growth curve. The correct option will show a curve that starts from the origin and approaches a horizontal line (asymptote) at \( V_0 \).

### Final Conclusion
Based on the analysis, the correct answer is option **A**, which represents the exponential curve of potential \( V \) against time \( t \) during the charging of a capacitor.

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