A fresh dry cell of 1.5 volt and two resistors of `10 k Omega` each are connected in series. An analog voltmeter measures a voltage of 0.5 volt across each of the resistor. A `100 mu F` capacitor is fully charged using the same source. The same voltmeter is now used to measure the voltage across It. The initial value of the current and the time in which the voltmeter reading falls to 0.5 volt are respectively.

Options

A. `60 muA,11s`
B. `120 muA,15 s`
C. `150 muA,m15 s`
D. `150 muA,11s`

Explanation

To solve the problem step by step, we will analyze the circuit and use the relevant formulas.

### Step 1: Understand the Circuit Configuration
We have a fresh dry cell of 1.5 volts connected in series with two resistors, each of 10 kΩ. The total resistance in the circuit is:
\[ R_{total} = R_1 + R_2 = 10 \, k\Omega + 10 \, k\Omega = 20 \, k\Omega \]

### Step 2: Calculate the Total Current in the Circuit
Using Ohm's law, the total current \( I \) in the circuit can be calculated as:
\[ I = \frac{V}{R_{total}} = \frac{1.5 \, V}{20 \, k\Omega} \]
Converting \( 20 \, k\Omega \) to ohms:
\[ 20 \, k\Omega = 20,000 \, \Omega \]
Now substituting the values:
\[ I = \frac{1.5}{20000} = 0.000075 \, A = 75 \, \mu A \]

### Step 3: Analyze the Voltmeter Reading
The voltmeter measures 0.5 volts across each resistor. This indicates that the voltmeter is also part of the circuit and has some internal resistance. The reading across each resistor confirms that the circuit is functioning correctly.

### Step 4: Charging the Capacitor
A capacitor of \( 100 \, \mu F \) is charged using the same source. When the capacitor is fully charged, it will have the same voltage as the source, which is 1.5 volts.

### Step 5: Determine the Initial Current through the Capacitor
When the capacitor is first connected, it behaves like a short circuit, and the initial current \( I_0 \) can be calculated using:
\[ I_0 = \frac{V}{R} \]
Where \( R \) is the total resistance in the circuit, which is still \( 20 \, k\Omega \):
\[ I_0 = \frac{1.5 \, V}{20 \, k\Omega} = 75 \, \mu A \]

### Step 6: Calculate the Time for the Voltage to Fall to 0.5 Volts
The voltage across a charging capacitor can be described by the equation:
\[ V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) \]
Where:
- \( V_0 = 1.5 \, V \) (initial voltage)
- \( V(t) = 0.5 \, V \) (voltage after time \( t \))
- \( R = 20 \, k\Omega = 20000 \, \Omega \)
- \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)

First, calculate \( RC \):
\[ RC = 20000 \, \Omega \times 100 \times 10^{-6} \, F = 2 \, seconds \]

Now, substituting into the voltage equation:
\[ 0.5 = 1.5 \left(1 - e^{-\frac{t}{2}}\right) \]
Rearranging gives:
\[ \frac{0.5}{1.5} = 1 - e^{-\frac{t}{2}} \]
\[ \frac{1}{3} = 1 - e^{-\frac{t}{2}} \]
\[ e^{-\frac{t}{2}} = 1 - \frac{1}{3} = \frac{2}{3} \]

Taking the natural logarithm:
\[ -\frac{t}{2} = \ln\left(\frac{2}{3}\right) \]
\[ t = -2 \ln\left(\frac{2}{3}\right) \]

Calculating \( t \):
\[ t \approx -2 \times (-0.4055) \approx 0.811 \, seconds \]

### Final Results
The initial current is \( 75 \, \mu A \) and the time for the voltmeter reading to fall to 0.5 volts is approximately \( 0.811 \, seconds \).