****Show all forces, motion/free-body diagrams, calculations, equations, significant figures and work. **** **Problem Statement:**A 1.5 μF capacitor and a 4.2 μF capacitor are connected in parallel across a 240 V potential difference. Calculate the total energy in joules stored in the capacitors.**Explanation:**When capacitors are connected in parallel, the total capacitance (C_total) is the sum of the individual capacitances:\[ C_{\text{total}} = C_1 + C_2 \]Given:- \( C_1 = 1.5 \, \mu \text{F} = 1.5 \times 10^{-6} \, \text{F} \)- \( C_2 = 4.2 \, \mu \text{F} = 4.2 \times 10^{-6} \, \text{F} \)- Voltage (V) = 240 VFirst, calculate the total capacitance:\[ C_{\text{total}} = 1.5 \times 10^{-6} + 4.2 \times 10^{-6} = 5.7 \times 10^{-6} \, \text{F} \]The energy \( E \) stored in a capacitor is given by the formula:\[ E = \frac{1}{2} C V^2 \]Substitute the total capacitance and voltage into the formula to find the energy:\[ E = \frac{1}{2} \times 5.7 \times 10^{-6} \times (240)^2 \]Calculate:\[ E = \frac{1}{2} \times 5.7 \times 10^{-6} \times 57600 \]\[ E = 0.163944 \, \text{J} \]Therefore, the total energy stored in the capacitors is approximately 0.164 joules.

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A 1.5 µF capacitor and a 4.2 µF capacitor are connected in parallel across a 240 V potential difference. Calculate the total energy in joules stored in the capacitors.