Question

Calculate the number of 2 µF capacitors must be connected in parallel to store a charge of XC (x=11) with a potential difference of 125 V across the capacitors?​

Answer 1

Answer:

Let, x capacitors be connected in parallel.

So, Ceq = 2x μF (Capacitance = C₁ + C₂ + ... in parallel and, 1/Capacitance = 1/C₁ + 1/C₂ + ... in series)

Given that V = 125 Volts

Charge 11 Coulombs =

So, Q = CV

Or, 11 = 2x x 10-6 × 125

Or, x = 44000

Answer by calculation is correct but it is impractical. So, if your answer do not match, then pls check the value of Q correctly in question. If it is some different value, then replace it in place of 11, and then solve. But the method is absolutely correct.

Explanation:

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