Multiple choice

Two capacitors of capacitance 6 μF and 12 μF are connected in series with a battery. What is the total battery voltage if voltage across the 6 μF capacitor is 2 V?

  1. 6.000012 V

  2. 12 x 10-12 V

  3. 3 V

  4. 2 V

  5. 24 x 10-12 V

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C Correct answer
Explanation

V = V1 + V2  Q = 6 x 10-6 X 2 = 12 μ C As C2 is in series, same amount of charge will flow through it. Now V2 = Q/C2 = (12 x 10-6)/(12 x 10-6) = 1 V   Total battery voltage, V = 2 + 1 = 3 V 

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