Question

The flash unit in a camera uses a 3.0 V battery to charge acapacitor. The capacitor is then discharged through a flashlamp.The discharge takes 7 µs, andthe average power dissipated in the flashlamp is 8 W. What is the capacitance of the capacitor?

Answer 1

The capacitance of the capacitor is approximately 18.67µF.

Step-by-step explanation:

To calculate the capacitance of the capacitor, we can use the formula:

C = Q/V, where C is capacitance, Q is charge, and V is voltage.

First, we need to find the charge stored in the capacitor. The power dissipated in the flashlamp can be calculated using the formula P = Q/t, where P is power, Q is charge, and t is time. Rearranging the formula, we have Q = P * t. Plugging in the given values, we get Q = 8W × 7µs = 56µC.

Next, we can use the formula C = Q/V to find the capacitance. Plugging in the given values, we have C = 56µC / 3.0V = 18.67µF. Therefore, the capacitance of the capacitor is approximately 18.67µF.

Answer 2

`C=2.54\cdot 10^5\ F`

Step-by-step explanation:

Energy Stored on a Capacitor

This energy is stored in the electric field present when a voltage V is applied to a capacitor C. It can be computed by the formula

`\displaystyle E=(CV^2)/(2)`

We can compute the energy by knowing the power dissipated in the flashlamp is P=8 W in a time of
`7 \mu s=7\cdot 10^(-6)\ sec`

`\displaystyle E=(P)/(t)=(8\ W)/(7\cdot 10^(-6)\ sec)=1.14\cdot 10^6\ J`

Solving the first equation for C

`\displaystyle C=(2E)/(V^2)`

`\displaystyle C=(2* 1.14\cdot 10^6)/(3^2)`

`\boxed{C=2.54\cdot 10^5\ F}`