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

Options

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

Explanation

To solve the problem of finding the time required for the capacitor to charge up to 0.75 V in a parallel combination of a resistor and a capacitor connected to a voltage source, we can follow these steps:

### Step 1: Understand the Circuit
We have a resistor (R = 0.1 MΩ = 100,000 Ω) and a capacitor (C = 10 µF = 10 × 10^-6 F) connected in parallel across a voltage source of 1.5 V.

### Step 2: Determine the Time Constant
For an RC circuit, the time constant (τ) is given by the formula:
\[ \tau = R \times C \]
Substituting the values:
\[ \tau = 100,000 \, \Omega \times 10 \times 10^{-6} \, F = 1 \, s \]

### Step 3: Use the Charging Formula
The voltage across the capacitor as it charges is given by:
\[ V(t) = V_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
Where:
- \(V(t)\) is the voltage across the capacitor at time \(t\),
- \(V_0\) is the final voltage (1.5 V),
- \(e\) is the base of the natural logarithm,
- \(\tau\) is the time constant.

We want to find the time \(t\) when \(V(t) = 0.75 V\).

### Step 4: Set Up the Equation
Substituting the known values into the equation:
\[ 0.75 = 1.5 \left(1 - e^{-\frac{t}{1}}\right) \]

### Step 5: Solve for \(e^{-\frac{t}{1}}\)
Rearranging the equation:
\[ \frac{0.75}{1.5} = 1 - e^{-t} \]
\[ 0.5 = 1 - e^{-t} \]
\[ e^{-t} = 0.5 \]

### Step 6: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ -t = \ln(0.5) \]
\[ t = -\ln(0.5) \]

### Step 7: Calculate \(t\)
Using the value of \(\ln(0.5) \approx -0.693\):
\[ t \approx -(-0.693) = 0.693 \, s \]

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

---