Question

The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 360 V , which charges a capacitor. The capacitor is then discharged through a flashlamp. The discharge takes 12 μs , and the average power dissipated in the flashlamp is 1.0×105 W .What is the capacitance of the capacitor? use 2 sigfigsI have tried 0.27 F and 0.26 F but so far both are wrong on mastering physics.

Answer 1

Answer:

The capacitance of the capacitor is $1.85*10^{-5}F$

Explanation:

To solve this exercise it is necessary to 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

Solving to find the Energy we have,

$E = P*t$

Our values are:

$P = 1*10^5W$

$t = 12*10^{-6}s$

Then,

$E= (1*10^5)(12*10^{-6})$

$E = 1.2J$

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

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

Solving for C=

$C = \frac{1}{2}\frac{E}{V^2}$

$C = 2\frac{1.2}{(360^2-3^2)}$

$C = 1.85*10^{-5}F$

Therefore the capacitance of the capacitor is $1.85*10^{-5}F$

Answer 2

The capacitance of the capacitor is $1.85 \times 10^{-5}F$

Calculation of capacitor capacitance:

We know that

$E = p\times t\\\\= (1\times 10^{5}) \times (12 \times 10^{-6})$

= 1.2J

Now

With the energy calculated above we can now determine the Capacitance in a capacitor via  the energy for capacitor equation.

Now the capacitance is

$E = \frac{1}{2} CV^2\\\\C = 2 \frac{1.2}{360^2 - 3^2} \\\\= 1.85 \times 10^{-5}F$

Here the above formula should be used for determining the capacitance.

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