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sdas [7]
4 years ago
7

(a) A 2.00-µF capacitor is connected to a 18.0-V battery. How much energy is stored in the capacitor?

Physics
2 answers:
marshall27 [118]4 years ago
8 0
<h2>Answer:</h2>

1.8 x 10⁻⁵J

<h2>Explanation:</h2>

The energy (E) stored in a capacitor of capacitance, C,  when a voltage, V, is supplied is given by;

E = \frac{1}{2} x C x V²              -------------------(i)

Now, from the question;

C = 2.00μF = 2.00 x 10⁻⁶F

V = 18.0V

Substitute these values into equation (i) as follows;

E = \frac{1}{2} x 2.00 x 10⁻⁶ x 18.0

E = 1.8 x 10⁻⁵J

Therefore, the quantity of energy stored in the capacitor is 1.8 x 10⁻⁵J

Sedaia [141]4 years ago
3 0

Answer:

1.8 x 10⁻⁵J.

Explanation:

E = 1/2 x C x V²

Where,

E = energy stored in a capacitor

C = capaciitance

V = Voltage

From the question, given:

C = 2.00μF

= 2.00 x 10⁻⁶F

V = 18.0V

E = 1/2 x 2.00 x 10⁻⁶ x 18.0

= 1.8 x 10⁻⁵J.

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