Three capacitors are connected in parallel to a 100 V battery as shown in figure. What is the total energy stored in the combination of capacitor?
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First of all we have to find out equivalent capacitance.
All capacitors are in parallel so, equivalent capacitance is Ceq
e.g., Ceq = C₁ + C₂ + C₃
C₁ = 30μF , C₂ = 20μF and C₃ = 10μF
∴ Ceq = 30μF + 20μF + 10μF
Ceq = 60 μF
Total energy stored in the combination of capacitor = heat energy + potential energy
= 1/2 CV² + 1/2 CV²
= CV²
Here, C = Ceq = 60μF and V = 100v
∴ total energy = 60 × 10⁻⁶ × 100 Joule
Total energy = 6000 × 10⁻³ J = 6 × 10⁻³ J
All capacitors are in parallel so, equivalent capacitance is Ceq
e.g., Ceq = C₁ + C₂ + C₃
C₁ = 30μF , C₂ = 20μF and C₃ = 10μF
∴ Ceq = 30μF + 20μF + 10μF
Ceq = 60 μF
Total energy stored in the combination of capacitor = heat energy + potential energy
= 1/2 CV² + 1/2 CV²
= CV²
Here, C = Ceq = 60μF and V = 100v
∴ total energy = 60 × 10⁻⁶ × 100 Joule
Total energy = 6000 × 10⁻³ J = 6 × 10⁻³ J
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