Three capacitors of capacitances `2 muF`, `3mu F` and `6muF` are connected in series with a 12 V battery. All the connecting wire are disconnected, the three positive plates are connected together and the three negative plates are connected together. Find the charges on the three capacitors after the reconnection.

Options

A. ` Q_1 = (7)/(11) muC, Q_2 =(10)/(11) muC and Q_3 = (21)/(11) muC`
B. ` Q_1 = (72)/(11) muC, Q_2 =(18/(11) muC and Q_3 = (216)/(11) muC`
C. ` Q_1 = (72)/(11) muC, Q_2 =(108)/(11) muC and Q_3 = (216)/(11) muC`
D. ` Q_1 = (2)/(11) muC, Q_2 =(108)/(11) muC and Q_3 = (216)/(11) muC`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the Equivalent Capacitance
When capacitors are connected in series, the equivalent capacitance \( C_{\text{eq}} \) can be calculated using the formula:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
where \( C_1 = 2 \mu F \), \( C_2 = 3 \mu F \), and \( C_3 = 6 \mu F \).

Substituting the values:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]
Finding a common denominator (which is 6):
\[ \frac{1}{C_{\text{eq}}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \]
Thus,
\[ C_{\text{eq}} = 1 \mu F \]

### Step 2: Calculate the Total Charge
The total charge \( Q \) stored in the equivalent capacitor when connected to a voltage \( V \) is given by:
\[ Q = C_{\text{eq}} \times V \]
Substituting \( C_{\text{eq}} = 1 \mu F \) and \( V = 12 V \):
\[ Q = 1 \mu F \times 12 V = 12 \mu C \]

### Step 3: Charge Distribution After Reconnection
After disconnecting the battery and connecting the positive plates together and the negative plates together, the charge will redistribute among the capacitors.

Let the charges on the capacitors be \( Q_1 \), \( Q_2 \), and \( Q_3 \) for \( C_1 \), \( C_2 \), and \( C_3 \) respectively. Since the capacitors are now in parallel, the total charge must equal the total charge calculated earlier:
\[ Q_1 + Q_2 + Q_3 = Q = 12 \mu C \]

### Step 4: Relate Charges to Voltages
For capacitors in parallel, the voltage across each capacitor is the same. Thus, we can express the charges in terms of the capacitances and a common voltage \( V_f \):
\[ Q_1 = C_1 \times V_f = 2 \mu F \times V_f \]
\[ Q_2 = C_2 \times V_f = 3 \mu F \times V_f \]
\[ Q_3 = C_3 \times V_f = 6 \mu F \times V_f \]

### Step 5: Substitute and Solve for \( V_f \)
Substituting these expressions into the charge conservation equation:
\[ 2V_f + 3V_f + 6V_f = 12 \mu C \]
\[ (2 + 3 + 6)V_f = 12 \mu C \]
\[ 11V_f = 12 \mu C \]
\[ V_f = \frac{12 \mu C}{11} \approx 1.09 \mu C \]

### Step 6: Calculate Individual Charges
Now substituting \( V_f \) back to find individual charges:
\[ Q_1 = 2 \mu F \times \frac{12}{11} \mu C = \frac{24}{11} \mu C \approx 2.18 \mu C \]
\[ Q_2 = 3 \mu F \times \frac{12}{11} \mu C = \frac{36}{11} \mu C \approx 3.27 \mu C \]
\[ Q_3 = 6 \mu F \times \frac{12}{11} \mu C = \frac{72}{11} \mu C \approx 6.55 \mu C \]

### Final Charges
Thus, the final charges on the capacitors are:
- \( Q_1 \approx 2.18 \mu C \)
- \( Q_2 \approx 3.27 \mu C \)
- \( Q_3 \approx 6.55 \mu C \)

---