A resistor 'R' and `2(mu)F` capacitor in series is connected through a switch to 200 V direct supply. A cross the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (`log_(10) 2.5 = 0.4`)

Options

A. `1.7xx10^(5) Omega`
B. `2.7xx10^(6) Omega`
C. `3.3xx10^(7) Omega`
D. `1.3xx10^(4) Omega`

Explanation

To solve the problem, we need to determine the value of the resistor \( R \) that will allow a neon bulb to light up after 5 seconds when connected in series with a 2 µF capacitor across a 200 V supply. The bulb lights up at 120 V.

### Step-by-Step Solution:

1. **Identify the formula for the voltage across the capacitor**:
The voltage \( V \) across a charging capacitor in an RC circuit is given by:
\[ V = V_0 \left(1 - e^{-\frac{t}{RC}}\right) \]
where:
- \( V_0 \) is the supply voltage (200 V),
- \( V \) is the voltage across the capacitor (120 V),
- \( t \) is the time (5 s),
- \( R \) is the resistance,
- \( C \) is the capacitance (2 µF = \( 2 \times 10^{-6} \) F).

2. **Set up the equation using the known values**:
Plugging in the values we know:
\[ 120 = 200 \left(1 - e^{-\frac{5}{R \cdot 2 \times 10^{-6}}}\right) \]

3. **Rearranging the equation**:
Divide both sides by 200:
\[ \frac{120}{200} = 1 - e^{-\frac{5}{R \cdot 2 \times 10^{-6}}} \]
Simplifying gives:
\[ 0.6 = 1 - e^{-\frac{5}{R \cdot 2 \times 10^{-6}}} \]
Rearranging further:
\[ e^{-\frac{5}{R \cdot 2 \times 10^{-6}}} = 1 - 0.6 = 0.4 \]

4. **Taking the natural logarithm**:
Taking the natural logarithm of both sides:
\[ -\frac{5}{R \cdot 2 \times 10^{-6}} = \ln(0.4) \]
Thus:
\[ \frac{5}{R \cdot 2 \times 10^{-6}} = -\ln(0.4) \]

5. **Calculating \(-\ln(0.4)\)**:
Using the property of logarithms:
\[ \ln(0.4) = \ln\left(\frac{2.5}{10}\right) = \ln(2.5) - \ln(10) \]
We know from the problem statement that:
\[ \log_{10}(2.5) = 0.4 \implies \ln(2.5) = 2.303 \times 0.4 = 0.9212 \]
Therefore:
\[ -\ln(0.4) = -\ln(2.5) + \ln(10) = -0.9212 + 2.303 = 1.3812 \]

6. **Substituting back to find \( R \)**:
Substitute \(-\ln(0.4)\) back into the equation:
\[ \frac{5}{R \cdot 2 \times 10^{-6}} = 1.3812 \]
Rearranging gives:
\[ R = \frac{5}{1.3812 \cdot 2 \times 10^{-6}} = \frac{5}{2.7624 \times 10^{-6}} \approx 1.81 \times 10^6 \, \Omega \]

7. **Final result**:
Therefore, the value of \( R \) is approximately:
\[ R \approx 1.81 \, M\Omega \]