Question

Two capacitor C1 = 5 µF and C = 3 µF are connected in parallel with a source of 12V. Calculate 2 the total energy stored in the capacitors

Answer 1

Given:

Two capacitors of c1=5uf and c2=3uf are connected in parallel with a source of 12V.

To find:

Total energy in the capacitors.

Calculation:

Since the capacitors are connected in parallel combination so both the capacitors will have a potential difference of 12 volts.

Total energy

$\sf{ = \dfrac{1}{2} (C1) {V}^{2} + \dfrac{1}{2} (C2) {V}^{2} }$

$\sf{ = \dfrac{1}{2} (C1 + C2) {V}^{2} }$

$\sf{ = \dfrac{1}{2} \{(5 + 3) \times {10}^{ - 6} \} \times {(12)}^{2} }$

$\sf{ = \dfrac{1}{2} \{8 \times {10}^{ - 6} \} \times144 }$

$\sf{ = \{8 \times {10}^{ - 6} \} \times72 }$

$\sf{ = 576 \times {10}^{ - 6} \: joule}$

So, final answer is:

$\boxed{ \red{ \large{ \sf{energy = 576 \times {10}^{ - 6} \: joule}}}}$