Question

Two capacitors of values of 1.0 μF and 0.50 μF are connected in parallel. The system is hooked up to a 100 V battery. Find the electrical potential energy stored in the 1.0 μF capacitor.Group of answer choices1.0x10-2 J1.9x10-3 J5.0x10-3 J6.5x10-3 J

Answer 1

5.0x10-3 J

Step-by-step explanation

Step 1

find the equivalent capacitor,When capacitors are connected together in parallel the total or equivalent capacitance, CT in the circuit is equal to the sum of all the individual capacitors added together

so

`\begin{gathered} C_(eq)=1.0\mu F+0.50\mu F \\ C_(eq)=1.50\mu F \end{gathered}`

Step 2

We can proceed to use this formula and solve for the potential energy stored in the 1.0mF capacitor.

`\begin{gathered} U=(1)/(2)cV^2 \\ U=(1)/(2)(0.000001c)(100^2) \\ U=0.005\text{ J} \\ U=5\cdot10^{-3\text{ }}J \end{gathered}`

therefore, the answer is

5.0x10-3 J

I hope this helps you