Capacitors `C_(1)=1 muF` and `C_(2)=2muF` are separately charged from the same battery. They are allowed to discharge separately through equal resistors

Options

A. The currents in the two dicharging circuits at `t=0` is zero
B. The currents in the two discharging circuits at `t=0` are equal but not zero
C. The currents in the two discharging circuits at `t=0` are unequal
D. `C_(1)` loses `50%` of its initial charge sooner than `C_(20` loses `50%` of its initial charge.

Explanation

To solve the problem step by step, we will analyze the discharging process of the capacitors \( C_1 \) and \( C_2 \) through equal resistors.

### Step 1: Write the equations for discharging current
The current \( I \) during the discharging of a capacitor is given by the equation:
\[ I(t) = I_0 e^{-t/(RC)} \]
where \( I_0 \) is the initial current, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is time.

For capacitor \( C_1 \):
\[ I_1(t) = I_{10} e^{-t/(R C_1)} \]
For capacitor \( C_2 \):
\[ I_2(t) = I_{20} e^{-t/(R C_2)} \]

### Step 2: Determine the initial currents
Since both capacitors are charged from the same battery and discharged through the same resistor, the initial currents \( I_{10} \) and \( I_{20} \) are equal:
\[ I_{10} = I_{20} = \frac{V}{R} \]
where \( V \) is the voltage of the battery.

### Step 3: Analyze the current at \( t = 0 \)
At \( t = 0 \):
\[ I_1(0) = I_{10} \quad \text{and} \quad I_2(0) = I_{20} \]
Thus, both currents are equal and non-zero:
\[ I_1(0) = I_2(0) = \frac{V}{R} \]

### Step 4: Determine the time for 50% discharge
The charge \( Q \) on a capacitor is given by:
\[ Q(t) = Q_0 e^{-t/(RC)} \]
For 50% discharge, we set \( Q(t) = \frac{Q_0}{2} \):
\[ \frac{Q_0}{2} = Q_0 e^{-t/(RC)} \]
Dividing both sides by \( Q_0 \) and rearranging gives:
\[ \frac{1}{2} = e^{-t/(RC)} \]
Taking the natural logarithm:
\[ \ln\left(\frac{1}{2}\right) = -\frac{t}{RC} \]
Thus,
\[ t = -RC \ln\left(\frac{1}{2}\right) = RC \ln(2) \]

For \( C_1 \):
\[ t_1 = R C_1 \ln(2) \]
For \( C_2 \):
\[ t_2 = R C_2 \ln(2) \]

### Step 5: Compare the discharge times
Given that \( C_1 = 1 \mu F \) and \( C_2 = 2 \mu F \):
\[ t_1 = R (1 \times 10^{-6}) \ln(2) \]
\[ t_2 = R (2 \times 10^{-6}) \ln(2) \]
Since \( C_2 > C_1 \), it follows that:
\[ t_2 > t_1 \]
This means \( C_1 \) discharges 50% of its charge sooner than \( C_2 \).

### Conclusion
- The currents at \( t = 0 \) are equal but not zero.
- Capacitor \( C_1 \) loses 50% of its initial charge sooner than capacitor \( C_2 \).

### Final Answers
1. The currents in the two discharging circuits at \( t = 0 \) are equal but not zero.
2. Capacitor \( C_1 \) loses 50% of its initial charge sooner than capacitor \( C_2 \).