A capacitor `C_1` of capacitance `1muF` and a capacitor `C_2` of capacitance `2muF` are separately charged by a common battery for a long time. The two capacitors are then separately discharged through equal resistors. Both the discharge circuits are connected at `t = 0`.

Options

A. The current in each of the two discharging circuits is zero at `t = 0`
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_2` loses 50% of its initial charge.

Answer

To solve the problem step by step, let's analyze the situation with the two capacitors, \( C_1 \) and \( C_2 \), and their discharging process through equal resistors.

### Step 1: Understand the Initial Conditions
Both capacitors \( C_1 \) (1 µF) and \( C_2 \) (2 µF) are charged by a common battery for a long time. This means that they reach a steady state where they hold their maximum charge.

### Step 2: Calculate Initial Charge on Each Capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \cdot V \]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

Let \( V \) be the voltage of the battery.

- For capacitor \( C_1 \):
\[ Q_1 = C_1 \cdot V = 1 \times 10^{-6} \, \text{F} \cdot V = 1 \mu C \cdot V \]

- For capacitor \( C_2 \):
\[ Q_2 = C_2 \cdot V = 2 \times 10^{-6} \, \text{F} \cdot V = 2 \mu C \cdot V \]

### Step 3: Discharge Through Resistors
Both capacitors are discharged through equal resistors \( R \). The initial current \( I_0 \) in each circuit at \( t = 0 \) can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
Since both capacitors are connected to the same battery and have the same resistance, the initial current will be the same for both circuits.

### Step 4: Analyze Current Over Time
The current \( I(t) \) in the discharge circuit as a function of time is given by:
\[ I(t) = I_0 e^{-t/(RC)} \]
where \( R \) is the resistance and \( C \) is the capacitance of the capacitor being discharged.

### Step 5: Calculate Time Constants
The time constant \( \tau \) for each capacitor is given by:
\[ \tau = R \cdot C \]

- For capacitor \( C_1 \):
\[ \tau_1 = R \cdot C_1 = R \cdot 1 \times 10^{-6} \, \text{s} \]

- For capacitor \( C_2 \):
\[ \tau_2 = R \cdot C_2 = R \cdot 2 \times 10^{-6} \, \text{s} \]

### Step 6: Compare Time Constants
Since \( \tau_1 < \tau_2 \), capacitor \( C_1 \) will discharge faster than capacitor \( C_2 \).

### Step 7: Determine When Each Capacitor Loses 50% Charge
The time taken for a capacitor to lose 50% of its charge can be calculated using:
\[ I(t) = I_0 e^{-t/\tau} \]
Setting \( I(t) = \frac{I_0}{2} \) gives:
\[ \frac{1}{2} = e^{-t/\tau} \implies t = \tau \ln(2) \]
Thus:
- For \( C_1 \):
\[ t_1 = \tau_1 \ln(2) \]

- For \( C_2 \):
\[ t_2 = \tau_2 \ln(2) \]

Since \( \tau_1 < \tau_2 \), it follows that \( t_1 < t_2 \). Therefore, \( C_1 \) loses 50% of its charge sooner than \( C_2 \).

### Final Conclusion
- The initial current in both circuits at \( t = 0 \) is equal and non-zero.
- Capacitor \( C_1 \) loses 50% of its initial charge sooner than capacitor \( C_2 \).