A capacitor of capacitance 5.00 is charged to 24.0 V , while a second capacitor of capacitance 6.0 is charged to 12.0 V . Determine the energy stored in each capacitor.

Correct answer: B. 1.44 mJ and 0.432 mJ The energy stored in a capacitor is given by the formula U = (1)/(2)CV^2 . For the first capacitor: U_1 = (1)/(2) C_1 V_1^2 U_1 = (1)/(2) × 5.00 × 10^ -6 × (24.0)^2 U_1 = 1440 × 10^ -6 J = 1.44 mJ For the second capacitor: U_2 = (1)/(2) C_2 V_2^2 U_2 = (1)/(2) × 6.0 × 10^ -6 × (12.0)^2 U_2 = 432 × 10^ -6 J = 0.432 mJ

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Capacitors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Capacitors

A capacitor of capacitance 5.00 is charged to 24.0 V , while a...

A capacitor of capacitance

5.00 \mu\text{F}

is charged to

24.0 \text{ V}

, while a second capacitor of capacitance

6.0 \mu\text{F}

is charged to

12.0 \text{ V}

. Determine the energy stored in each capacitor.

Options

A $0.72 \text{ mJ}$ and $0.216 \text{ mJ}$
B $1.44 \text{ mJ}$ and $0.432 \text{ mJ}$
C $2.88 \text{ mJ}$ and $0.864 \text{ mJ}$
D $1.20 \text{ mJ}$ and $0.720 \text{ mJ}$

Correct answer

B. $1.44 \text{ mJ}$ and $0.432 \text{ mJ}$

Step-by-step solution

The energy stored in a capacitor is given by the formula

U = \dfrac{1}{2}CV^2

.

For the first capacitor:

U_1 = \dfrac{1}{2} C_1 V_1^2

U_1 = \dfrac{1}{2} \times 5.00 \times 10^{-6} \times (24.0)^2

U_1 = 1440 \times 10^{-6} \text{ J} = 1.44 \text{ mJ}

For the second capacitor:

U_2 = \dfrac{1}{2} C_2 V_2^2

U_2 = \dfrac{1}{2} \times 6.0 \times 10^{-6} \times (12.0)^2

U_2 = 432 \times 10^{-6} \text{ J} = 0.432 \text{ mJ}

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