Suppose we have two capacitors connected in series with a 10 V battery.What is the energy stored in the circuit if the capacitance of one is C₁ = 50 mF and the other is C₂ = 100 mF?

Question

Quizplus · Accepted Answer

The total capacitance of the circuit in series can be found using the formula:

1/C = 1/C1+ 1/C2

Plugging in the values,

1/C = 1/50 x 10^-3 + 1/100 x 10^-3

1/C = 1/33.33 x 10^-3

C = 30 x 10^-6 F

The charge stored in the circuit can be found using:

Q = CV

Q = (30 x 10^-6 F) x (10 V)

Q = 300 x 10^-6 C

The energy stored in the circuit can be found using:

E = 0.5 x CV^2

E = 0.5 x 30 x 10^-6 F x (10 V)^2

E = 1.5 x 10^-3 J

Therefore, the energy stored in the circuit is 1.67 J (rounded to 2 decimal places). The correct answer is C.

Suppose We Have Two Capacitors Connected in Series with a 10