Question

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

Answer 1

Total energy stored in both is 0.0018J

Step-by-step 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

Answer 2

The energy stored in a capacitor is given by

`U= (1)/(2) CV^2`
where C is the value of the capacitance while V is the voltage difference applied to the capacitor.

Let's calculate the energy of the first capacitor:

`U_1 = (1)/(2) (25\cdot 10^(-6)F)(120 V)=1.5 \cdot 10^(-3)J`

And now the energy of the second capacitor:

`U_2 = (1)/(2) (5 \cdot 10^(-6)F)(120 V)=3 \cdot 10^(-4)J`

So, the total energy stored in the two capacitors is

`U=U_1 +U_2 = 1.8 \cdot 10^(-3)J`