What is the energy stored in a capacitor of capacitance 4μF charged to 12V? (202

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Question: What is the energy stored in a capacitor of capacitance 4μF charged to 12V? (2022)

Options:

Correct Answer: 48μJ

Exam Year: 2022

Solution:

The energy stored (U) is given by U = 1/2 CV². Thus, U = 1/2 * 4μF * (12V)² = 288μJ.

What is the energy stored in a capacitor of capacitance 4μF charged to 12V? (2022)

What is the energy stored in a capacitor of capacitance 4μF charged to 12V? (2022)

Step 1: Identify the formula for the energy stored in a capacitor, which is U = 1/2 CV².
Step 2: Identify the values needed for the formula: capacitance (C) is 4μF and voltage (V) is 12V.
Step 3: Convert the capacitance from microfarads to farads: 4μF = 4 x 10^-6 F.
Step 4: Substitute the values into the formula: U = 1/2 * (4 x 10^-6 F) * (12V)².
Step 5: Calculate (12V)², which is 144V².
Step 6: Multiply 4 x 10^-6 F by 144V² to get 5.76 x 10^-4 J.
Step 7: Multiply by 1/2: U = 1/2 * 5.76 x 10^-4 J = 2.88 x 10^-4 J.
Step 8: Convert joules to microjoules: 2.88 x 10^-4 J = 288μJ.

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