A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge

Analysis

To solve this problem, we need to understand how capacitors behave when connected in series and how to calculate the total charge stored in the circuit.

Concept

1. Capacitors in Series: When capacitors are connected in series, the total (equivalent) capacitance ($C_T$) is calculated using the formula: $$\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2}$$Alternatively, for two capacitors:$C_T = \frac{C_1 \times C_2}{C_1 + C_2}$
2. Charge Formula: The total charge ($Q$) in a capacitor circuit is given by: $$Q = C_T \times V$$Where$V$ is the supply voltage. In a series circuit, the charge on each capacitor is the same as the total charge supplied by the source.

Step-by-Step Solution

Step 1: Calculate the equivalent capacitance ($C_T$) Given: $C_1 = 3\mu F$ and $C_2 = 6\mu F$

Convert microfarads to farads: $2\mu F = 2 \times 10^{-6} F$

Step 2: Calculate the total charge ($Q$) Given: $V = 400V$

Step 3: Express in standard form

Conclusion

Comparing our result to the options provided, the calculated charge is $8 \times 10^{-4} C$. This matches option A.

Exam Tip

When capacitors are in series, the equivalent capacitance is always smaller than the smallest individual capacitor in the circuit. If you ever calculate a total capacitance larger than your smallest capacitor, double-check your math! Also, remember that in series, the charge ($Q$) is constant across all capacitors, but the voltage ($V$) is divided.