Question

Two capacitors of capacitances \( 1\mu F \) and \( 2\mu F \) can separately withstand potentials of \( 6 \) kV and \( 4 \) kV respectively. The total potential, they together can withstand when they are connected in series is: 

• A. \( 9 \) kV
• B. \( 3 \) kV
• C. \( 6 \) kV
• D. \( 2 \) kV
  \

Hint:

In a series connection of capacitors, the charge remains the same across each capacitor, and the total voltage is the sum of the individual withstand voltages.

Answer 1

The Correct Option is A

Step 1: Understanding Series Connection of Capacitors 
When capacitors are connected in series, the charge on each capacitor is the same. The total voltage across the series combination is the sum of the individual voltages: \[ V_{\text{total}} = V_1 + V_2. \] Given: - Capacitance \( C_1 = 1 \mu F \), withstand voltage \( V_1 = 6 \) kV. - Capacitance \( C_2 = 2 \mu F \), withstand voltage \( V_2 = 4 \) kV. 

Step 2: Relationship Between Charge and Voltage 
Since the charge remains the same in series, \[ Q = C_1 V_1 = C_2 V_2. \] Substituting values, \[ (1 \times 6) = (2 \times 4). \] \[ Q = 6 \mu C = 8 \mu C. \] 

Step 3: Finding the Total Potential 
Total potential: \[ V_{\text{total}} = V_1 + V_2 = 6 + 3 = 9 \text{ kV}. \] 

Step 4: Conclusion 
Thus, the capacitors together can withstand a total voltage of: \[ \boxed{9 \text{ kV}}. \]