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. $0.5\times10^{-3}\,J$
• D. $5.0\times10^{-2}\,J$

Hint:

For capacitors in parallel, capacitances add directly: $C_{eq}=C_1+C_2+\dots$. The energy stored in a capacitor is given by $U=\frac{1}{2}CV^2$.

Answer 1

The Correct Option is D

Step 1: Determine the equivalent capacitance.
When capacitors are connected in parallel, their capacitances add directly. Therefore, \[ C_{eq} = C_1 + C_2 + C_3 + \cdots + C_{10} \] Each capacitor has capacitance $1\,\mu F = 1\times10^{-6}\,F$. Thus, \[ C_{eq} = 10 \times 1\times10^{-6} = 10^{-5}\,F \]
Step 2: Use the energy stored formula.
The energy stored in a capacitor is given by \[ U = \frac{1}{2} C V^2 \] Substituting the values: \[ U = \frac{1}{2} \times 10^{-5} \times (100)^2 \]
Step 3: Perform the calculation.
\[ (100)^2 = 10^4 \] \[ U = \frac{1}{2} \times 10^{-5} \times 10^4 \] \[ U = \frac{1}{2} \times 10^{-1} \] \[ U = 0.5 \times 10^{-1} = 5 \times 10^{-2}\,J \]
Step 4: Compare with the options.
The calculated energy stored in the system is \[ U = 5\times10^{-2}\,J \] which corresponds to option (D).
Final Answer: $5\times10^{-2}\,J$.