A 15-$\mu$F capacitor and a 25-$\mu$F capacitor are connected in parallel,and charged to a potential difference of 60 V.How much energy is then stored in this capacitor combination?
A)50 mJ
B)18 mJ
C)32 mJ
D)72 mJ
E)45 mJ

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by the formula: 
E = 1/2 * C * V^2
where E is energy stored, C is capacitance, and V is potential difference.
So, for the given capacitors in parallel, the equivalent capacitance will be:
C_eq = C1 + C2
C_eq = 15 μF + 25 μF
C_eq = 40 μF
Now, using the formula, the total energy stored will be:
E = 1/2 * C_eq * V^2
E = 1/2 * 40 μF * (60 V)^2
E = 72 mJ
Therefore, the answer is D) 72 mJ.

A 15- μ\muμ F Capacitor and a 25- μ\muμ F Capacitor Are Connected in Parallel,and Charged to a Potential

A 15- $\mu$ F Capacitor and a 25- $\mu$ F Capacitor Are Connected in Parallel,and Charged to a Potential