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

Answer:

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

Answer 2

The energy stored in a capacitor is given by
$U= \frac{1}{2} CV^2$
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:
$U_1 = \frac{1}{2} (25\cdot 10^{-6}F)(120 V)=1.5 \cdot 10^{-3}J$

And now the energy of the second capacitor:
$U_2 = \frac{1}{2} (5 \cdot 10^{-6}F)(120 V)=3 \cdot 10^{-4}J$

So, the total energy stored in the two capacitors is
$U=U_1 +U_2 = 1.8 \cdot 10^{-3}J$