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3 years ago

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The flash unit in a camera uses a 3.0 V battery to charge acapacitor. The capacitor is then discharged through a flashlamp.The d

ischarge takes 7 µs, andthe average power dissipated in the flashlamp is 8 W. What is the capacitance of the capacitor?

1 answer:

Varvara68 [4.7K]3 years ago

6 0

Answer:

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

Explanation:

<u>Energy Stored on a Capacitor</u>

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

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

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

$\displaystyle E=\frac{P}{t}=\frac{8\ W}{7\cdot 10^{-6}\ sec}=1.14\cdot 10^6\ J$

Solving the first equation for C

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

$\displaystyle C=\frac{2\times 1.14\cdot 10^6}{3^2}$

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

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Explanation:

Given that,

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Distance = 5.50 cm

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$\Delta U=-W$

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Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

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Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

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Read 2 more answers

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Marina CMI [18]

Answer:

(a) $\vec{J}=176.4\frac{kg*m}{s} \hat{j}$

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(b) The average force is:

$F=\frac{J}{\Delta t}\\F=\frac{176.4\frac{kg*m}{s}}{0.083s}\\F=2125.30N$

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