Question

Ten capacitors, each of capacitance 1 \(\mu\)F, are connected in parallel to a source of 100 V. The total energy stored in the system is equal to:

• A. \(10^{-2}\) J
• B. \(10^{-3}\) J
• C. \(0.5 \times 10^{-3}\) J
• D. \(5.0 \times 10^{-2}\) J

Hint:

For capacitors in parallel, the total energy stored is the sum of the energies stored in each individual capacitor.

Answer 1

The Correct Option is D

The energy stored in a single capacitor is given by the formula: \[ E = \frac{1}{2} C V^2 \] Where: - \(E\) is the energy stored, - \(C\) is the capacitance, - \(V\) is the voltage across the capacitor. For one capacitor, with \(C = 1 \mu F = 1 \times 10^{-6}\) F and \(V = 100\) V: \[ E_1 = \frac{1}{2} (1 \times 10^{-6}) (100)^2 = \frac{1}{2} \times 10^{-6} \times 10^4 = 5 \times 10^{-2} \, \text{J} \] Since the capacitors are connected in parallel, the total energy stored in the system is the sum of the energies of all the capacitors. There are 10 capacitors, so the total energy is: \[ E_{\text{total}} = 10 \times 5 \times 10^{-2} = 5.0 \times 10^{-2} \, \text{J} \] Thus, the correct answer is: \[ \text{(D) } 5.0 \times 10^{-2} \, \text{J}. \]