Question

Two capacitor C1 = 5 µF and C2 = 3 µF are connected in parallel with a source of 12V. Calculate the total energy stored in the capacitors

Answer 1

GiveN :

• C1 = 5 µF = 5 × 10^-6 F
• C2 = 3 µF = 3 × 10^-6F
• V = 12 C

To FinD :

• Energy stored in Capacitors

SolutioN :

If Capacitors are connected in parallel then,

⇒C_{eq} = C1 + C2

⇒ C_{eq} = (5 × 10^-6) + (3 × 10^-6)

⇒ C_{eq} = 8 × 10^-6 F

________________________

Use formula for Energy of Capacitor :

⇒U = ½CV²

⇒ U = ½ * (8 * 10^-6) (12)²

⇒ U = 4 × 10^-6 * 144

⇒ U = 576 × 10^-6

➭ Energy stored in Capacitors is 576 × 10^-6 J.

Answer 2

Given ,

• Two capacitor of capacitance 5 µF and 3 µF are connected in parrallel with source of 12 v battery

We know that , the equivalent capacitance in parrallel combination is given by

C = C' + C'' + .....

Thus ,

C = 5 × 10^(-6) + 3 × 10^(-6)

C = 10^(-6) { 5 + 3 }

C = 8 × 10^(-6) Farad

Now , the energy stored in a capacitor is given by

$\boxed{ \sf{U = \frac{1}{2} C(v)²}}$

Thus ,

U = 1/2 × 8 × 10^(-6) × (12)²

U = 4 × 144 × 10^(-6)

U = 576 × 10^(-6) Joule

Therefore ,

• The energy stored in a capacitor is 576 × 10^(-3) Joule