A 24 V battery charges a 250 µF capacitor through a 100 Ω resistor until the capacitor is full…

Step 1. The charge. $Q = CV = 250 \times 10^{-6}\ \text{F} \times 24\ \text{V} = 6.0 \times 10^{-3}\ \text{C} = 6.0\ \text{mC}$.

Step 2. The energy stored. $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 250 \times 10^{-6}\ \text{F} \times (24\ \text{V})^2 = 0.072\ \text{J}$.

Step 3. The battery. $W = QV = 6.0 \times 10^{-3}\ \text{C} \times 24\ \text{V} = 0.144\ \text{J}$.

Step 4. The heat. $0.144\ \text{J} - 0.072\ \text{J} = 0.072\ \text{J}$. The heat equals the stored energy.

Step 5. A different resistor. The heat is $0.072\ \text{J}$ again. The resistance changes how fast the capacitor charges. It does not change how much heat is made.

Step 6. The time constant. $RC = 100\ \Omega \times 250 \times 10^{-6}\ \text{F} = 0.025\ \text{s} = 25\ \text{ms}$.