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 step by step, we follow these calculations:

### Step 1: Understand the circuit
We have a resistor \( R \) and a capacitor \( C = 2 \, \mu F \) in series connected to a 200 V supply. The neon bulb lights up at a voltage of 120 V across the capacitor.

### Step 2: Write the voltage across the capacitor
The voltage \( V_C \) across the capacitor at any time \( t \) after closing the switch can be given by the formula:
\[ V_C(t) = V(1 - e^{-t/\tau}) \]
where \( V \) is the supply voltage (200 V) and \( \tau \) is the time constant defined as \( \tau = R \cdot C \).

### Step 3: Set up the equation for when the bulb lights up
We want the voltage across the capacitor to be 120 V when \( t = 5 \, s \):
\[ 120 = 200(1 - e^{-5/\tau}) \]

### Step 4: Solve for \( e^{-5/\tau} \)
Rearranging the equation:
\[ 1 - e^{-5/\tau} = \frac{120}{200} = 0.6 \]
Thus,
\[ e^{-5/\tau} = 1 - 0.6 = 0.4 \]

### Step 5: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ -\frac{5}{\tau} = \ln(0.4) \]
This gives us:
\[ \frac{5}{\tau} = -\ln(0.4) \]

### Step 6: Use the logarithmic identity
Using the property of logarithms, we can express \( \ln(0.4) \):
\[ \ln(0.4) = \ln\left(\frac{2.5}{10}\right) = \ln(2.5) - \ln(10) \]
Given \( \log_{10}(2.5) = 0.4 \), we can convert this to natural logarithm:
\[ \ln(2.5) \approx 0.4 \cdot 2.303 = 0.9212 \quad (\text{since } \ln(10) \approx 2.303) \]
Thus,
\[ \ln(0.4) \approx 0.9212 - 2.303 \approx -1.381 \]

### Step 7: Substitute back to find \( \tau \)
Substituting back:
\[ \frac{5}{\tau} = 1.381 \implies \tau = \frac{5}{1.381} \approx 3.63 \, s \]

### Step 8: Relate \( \tau \) to \( R \) and \( C \)
We know that:
\[ \tau = R \cdot C \]
Substituting \( C = 2 \times 10^{-6} \, F \):
\[ 3.63 = R \cdot (2 \times 10^{-6}) \]

### Step 9: Solve for \( R \)
Rearranging gives:
\[ R = \frac{3.63}{2 \times 10^{-6}} = 1.815 \times 10^{6} \, \Omega \]

### Final Answer
Thus, the value of \( R \) is approximately:
\[ R \approx 1.82 \, M\Omega \]