A parallel combination of 0.1 `M Omega` resistor and a `10muF` capacitor is connected across a 1.5 V source of negligible resistance. The time (in sec) required for the capacitor to get charged upto 0.75 V is approximately

Options

A. Infinite
B. `log_(e)2`
C. `log_(10)2`
D. zero

Explanation

To solve the problem, we need to determine the time required for a capacitor to charge up to a certain voltage when connected in parallel with a resistor across a voltage source.

### Step-by-Step Solution:

1. **Identify the Components**:
We have a resistor (R) of 0.1 MΩ (which is 0.1 x 10^6 Ω = 100,000 Ω) and a capacitor (C) of 10 µF (which is 10 x 10^-6 F = 0.00001 F) connected in parallel across a 1.5 V source.

2. **Understand the Charging Equation**:
The voltage across a charging capacitor in a circuit can be described by the equation:
\[ V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) \]
where:
- \( V(t) \) is the voltage across the capacitor at time \( t \),
- \( V_0 \) is the final voltage (1.5 V in this case),
- \( R \) is the resistance,
- \( C \) is the capacitance,
- \( e \) is the base of the natural logarithm.

3. **Calculate the Time Constant (τ)**:
The time constant \( \tau \) is given by:
\[ \tau = RC \]
Substituting the values:
\[ R = 0.1 \times 10^6 \, \Omega = 100,000 \, \Omega \]
\[ C = 10 \times 10^{-6} \, F = 0.00001 \, F \]
Therefore:
\[ \tau = 100,000 \times 0.00001 = 1 \, sec \]

4. **Set Up the Equation for Charging to 0.75 V**:
We want to find the time \( t \) when the voltage across the capacitor \( V(t) \) is 0.75 V:
\[ 0.75 = 1.5 \left(1 - e^{-\frac{t}{1}}\right) \]
Simplifying this:
\[ 0.75 = 1.5 - 1.5 e^{-t} \]
Rearranging gives:
\[ 1.5 e^{-t} = 1.5 - 0.75 = 0.75 \]
Dividing both sides by 1.5:
\[ e^{-t} = \frac{0.75}{1.5} = 0.5 \]

5. **Solve for \( t \)**:
Taking the natural logarithm of both sides:
\[ -t = \ln(0.5) \]
Therefore:
\[ t = -\ln(0.5) \approx 0.693 \, sec \]

### Final Answer:
The time required for the capacitor to charge up to 0.75 V is approximately **0.693 seconds**.