Question

The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 290 V, which charges a capacitor. The capacitor is then discharged through a flashlamp. The discharge takes 10 μs, and the average power dissipated in the flashlamp is 1.0×105 W. What is the capacitance of the capacitator?

Answer 1

Answer:

Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F

Explanation:

We apply the concepts related to Power and energy stored in a capacitor.

By definition we know that power is represented as

$P = \frac{E}{t}$

Where,

E= Energy

t = time

to find the Energy we have,

$E = P*t$

$P = 1*10^5Wt = 10*10^{-6}s$

$E= (1*10^5)(10*10^{-6})E = 1.0J$

With the energy found we can know calculate the Capacitance in a capacitor through the energy for capacitor equation, that is

$E=\frac{1}{2}CV^2$

$C = \frac{2E}{V^2}\\\\\\\frac{2\times 1 }{290^2 - 3^2\\} \\= 2.38 \times 10^-^5F$

Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F

Answer 2

Answer:

24 μF.

Explanation:

Given:

Time, t = 10 μs

P = 1.0×105 W

V1 = 3 V

V2 = 290 V

Power, P = Energy/time

Energy = 1/2 × C × V^2

1 × 10-5 × 1 × 10^-5 = 1/2 × C × (290^2 - 3^2)

C = 2/84091

= 2.38 × 10^-5 F

= 24 μF