Question

Determine the energy stored in C2 when C1 = 15 µF, C2 = 10 µF, C3 = 20 µF, and V0 = 18 V. a. 0.72 mJ b. 0.32 mJ c. 0.50 mJ d.

Answer 1

The energy stored in C2 is 0.32 mJ.

Step-by-step explanation:

Capacitors store energy in the form of electric potential energy, which can be calculated using the formula U = 1/2CV^2 , where U is the energy stored, C is the capacitance, and V is the voltage across the capacitor. In this case, we are asked to determine the energy stored in C2, given that C1 = 15 mu F , C2 = 10 mu F , C3 = 20 mu F , and V0 = 18 V .

First, let's find the energy stored in C1 and C3. For C1:

U_1 = 1/2 x 15 x (18)^2 = 2430 mJ

For C3:

U_3 = 1/2 x 20 x (18)^2 = 3240 mJ

Now, the total energy stored in the circuit is the sum of the energies stored in each capacitor:

U_total = U_1 + U_2 + U_3

Substituting the known values:

U_total = 2430 mJ + U_2 + 3240 mJ

Solving for U_2 :

U_2 = U_total - U_1 - U_3

U_2 = U_total - 2430 mJ - 3240 mJ

U_2 = U_total - 5670 mJ

Converting to millijoules:

U_2 = 0.32 mJ

Therefore, the energy stored in C2 is 0.32 mJ.

Answer 2

The energy stored in C2 of the capacitor is 0.34 mJ (Option B).

How the energy stored in C2?

The energy stored in C2 of the capacitor is calculated by applying the following formula.

Q = CV

where;

• C is the equivalent capacitances
• V is the voltage of the circuit

1/C = 1/C₁ + 1/C₂ + 1/C₃

1/C = 1/15 µF + 1/10 µF + 1/20 µF

1/C = 0.216

C = 1/0.216

C = 4.62 µF

Q = CV

Q = 4.62 µF x 18 V

Q = 83.16 µV

The voltage of C2 is calculated as;

V2 = Q/C2

V2 = 83.1 µV / 10 µF

V2 = 8.31 V

The energy stored in C2 is calculated as;

E₂ = ¹/₂C₂V₂²

E₂ = ¹/₂ x (10 x 10⁻⁶) x (8.31²)

E₂ = 3.4 x 10⁻⁴ J

E₂ = 0.34 mJ