The image contains a set of multiple-choice capacitor-related problems in Bengali. I'll walk you through solving a few of them.

Problem 205: Two capacitors with capacitances of 4.4 µF and 8 µF are connected in parallel with a 100V battery. What is the total charge stored by the capacitors?

Solution: In a parallel combination, the total charge $Q_{\text{total}}$ is given by: $Q_{\text{total}} = C_{\text{total}} \times V$ Where $C_{\text{total}} = C_1 + C_2$ (since it's a parallel circuit).

1. $C_1 = 4.4 \, \mu F = 4.4 \times 10^{-6} \, F$
2. $C_2 = 8 \, \mu F = 8 \times 10^{-6} \, F$
3. $V = 100 \, V$

Now, $C_{\text{total}} = 4.4 \times 10^{-6} + 8 \times 10^{-6} = 12.4 \times 10^{-6} \, F$ Then, the total charge $Q_{\text{total}}$: $Q_{\text{total}} = 12.4 \times 10^{-6} \times 100 = 12.4 \times 10^{-4} \, C = 1.24 \times 10^{-3} \, C = 12.4 \, mC$

So, the closest answer seems to match option (3).

Would you like to proceed with more problems or need further explanation?

Here are five related questions for further exploration:

1. How do capacitors behave differently when arranged in series versus parallel?
2. What is the formula for calculating the equivalent capacitance in series?
3. How does the stored energy in a capacitor change with voltage?
4. Why is the total charge different in series versus parallel connections?
5. What is the physical meaning of capacitance?

Tip: Remember that the capacitance in parallel adds up, whereas in series, the reciprocal sum determines the total capacitance.