An uncharged capacitor of capacitance C is connected with a battery of EMF V and a resistance R. The switch is closed at `t=0`. The time instant at which the current in the circuit is `(V)/(4R)` is t=

Options

A. 2RC
B. 3RC
C. `2RClog_(e)2`
D. `3RClog_(e)2`

Explanation

To solve the problem step-by-step, we will analyze the charging of a capacitor in an RC circuit and derive the time at which the current becomes \( \frac{V}{4R} \).

### Step 1: Understand the Circuit
When the switch is closed at \( t = 0 \), the capacitor starts charging through the resistor. The current in the circuit decreases over time as the capacitor charges.

### Step 2: Write the Charging Equation of the Capacitor
The charge \( Q \) on the capacitor at any time \( t \) is given by the equation:
\[ Q(t) = Q_0 (1 - e^{-t/RC}) \]
where \( Q_0 \) is the maximum charge on the capacitor, \( R \) is the resistance, and \( C \) is the capacitance.

### Step 3: Find the Expression for Current
The current \( I \) in the circuit is the rate of change of charge with respect to time:
\[ I = \frac{dQ}{dt} \]
Differentiating the charging equation:
\[ I = \frac{d}{dt} \left[ Q_0 (1 - e^{-t/RC}) \right] = Q_0 \cdot \frac{1}{RC} e^{-t/RC} \]
Substituting \( Q_0 = CV \) (maximum charge when fully charged):
\[ I = \frac{CV}{RC} e^{-t/RC} = \frac{V}{R} e^{-t/RC} \]

### Step 4: Set the Current Equal to \( \frac{V}{4R} \)
We need to find the time \( t \) when the current \( I \) is equal to \( \frac{V}{4R} \):
\[ \frac{V}{R} e^{-t/RC} = \frac{V}{4R} \]
Canceling \( \frac{V}{R} \) from both sides (assuming \( V \neq 0 \)):
\[ e^{-t/RC} = \frac{1}{4} \]

### Step 5: Solve for \( t \)
Taking the natural logarithm of both sides:
\[ -t/RC = \ln\left(\frac{1}{4}\right) \]
This can be rewritten as:
\[ -t/RC = -\ln(4) \]
Thus,
\[ t = RC \ln(4) \]

### Step 6: Simplify the Expression
Using the property of logarithms, we can express \( \ln(4) \) as \( 2\ln(2) \):
\[ t = 2RC \ln(2) \]

### Final Answer
The time instant at which the current in the circuit is \( \frac{V}{4R} \) is:
\[ t = 2RC \ln(2) \]