Two devices with capacitances of 25 μf and 5.0 μf are each charged with separate 120 v power supplies. calculate the total energy stored in the two capacitors. your answer submit

Total energy stored in both is 0.0018J

Explanation:

Formula to calculate the total energy stored between two capacitors = 1/2CV²

Energy stored the First Capacitor = 1/2 × 25× 10⁻⁶f × (120)²

Energy stored = 0.0015J

Energy stored in the second Capacitor = 1/2 × 5× 10⁻⁶f × (120)²

Energy stored = 0.0003J

Total Energy stored = Energy stored in capacitance 1 + Energy stored in Capacitor 2

= 0.0015J+ 0.0003J

= 0.0018J

i think if it were me i whould divide 300 dide 15

step-by-step explanation: