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How much energy must a 24-V battery expend to charge a 4-μF and a 6-μF capacitor fully when they are placed in parallel?

Answer :

In this case, the 24-V battery must expend 2880 μJ of energy to charge the 4-μF and 6-μF capacitors fully when they are placed in parallel.

To calculate the energy expended by a 24-V battery to charge a 4-μF and a 6-μF capacitor fully when they are placed in parallel, we can follow these steps:

1. First, determine the total capacitance (C_total) when the capacitors are in parallel:

C_total = C1 + C2

C_total = 4 μF + 6 μF

C_total = 10 μF

2. Calculate the total charge (Q) stored in the capacitors when fully charged:

Q = C_total * V

Q = 10 μF * 24 V

Q = 240 μC

3. The energy (E) stored in a capacitor is given by the formula:

E = (1/2) * C * V²

4. Calculate the energy stored in each capacitor:

For the 4-μF capacitor:

E1 = (1/2) *4 μF *(24 V)²

E1 = (1/2) * 4 * 576

E1 = 1152 μJ

For the 6-μF capacitor:

E2 = (1/2) * 6 μF * (24 V)²

E2 = (1/2) * 6 * 576

E2 = 1728 μJ

5. To find the total energy expended by the battery to charge both capacitors fully:

E_total = E1 + E2

E_total = 1152 μJ + 1728 μJ

E_total = 2880 μJ

Therefore, the 24-V battery must expend 2880 μJ of energy to charge the 4-μF and 6-μF capacitors fully when they are placed in parallel.

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