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.
