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Chapter 10: Problem 25

A camera flash releases \(3 \mathrm{~J}\) of energy, which is supplied by a capacitor. If the voltage of the flash unit is \(200 \mathrm{~V}\), the capacitance of the capacitor is (A) \(200 \mathrm{~F}\) (B) \(200 \mu \mathrm{F}\) (C) \(150 \mu \mathrm{F}\) (D) \(10^{-8} \mathrm{~F}\) (E) \(150 \mathrm{nF}\)

Short Answer

Expert verified
(A) 100 μF, (B) 120 μF, (C) 150 μF, (D) 180 μF. Answer: (C) 150 μF.

Step by step solution

01

Understand the formula for energy stored in a capacitor

The formula for the energy stored in a capacitor is given by: \(E = \frac{1}{2}CV^2\), where \(E\) is the energy, \(C\) is the capacitance, and \(V\) is the voltage.
02

Rearrange the formula to solve for capacitance

We want to find the capacitance \(C\), so we need to rearrange the formula to make \(C\) the subject. To do this, we can multiply both sides by 2 and then divide by \(V^2\). This gives us the formula: \(C = \frac{2E}{V^2}\).
03

Plug in the given values and solve for capacitance

Now, we can plug in the given values for energy (\(3 \mathrm{~J}\)) and voltage (\(200 \mathrm{~V}\)) into the formula: \(C = \frac{2(3 \mathrm{~J})}{(200 \mathrm{~V})^2}\).
04

Calculate and find the correct answer

Carrying out the calculation, we get: \(C = \frac{6}{40000} \mathrm{F} = 0.00015 \mathrm{F}\). Converting this to microfarads, we have: \(C = 150 \mathrm{\mu F}\). The correct answer is (C) \(150 \mu \mathrm{F}\).

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Most popular questions from this chapter

A battery with voltage \(V\) sends a current \(I\) through a resistor \(R\). If \(V\) is doubled, the power dissipated by the resistor (A) stays the same. (B) quadruples. (C) halves. (D) doubles. (E) goes down by a factor of four.

A battery \(V\) sends a current \(I\) through two resistors in series, \(R\) and \(2 R\). If the first resistor is doubled to \(2 R\), the current \(I\) is now (A) the same as it was initially. (B) half what it was initially. (C) three-quarters what it was initially. (D) four-thirds what it was initially. (E) twice what is was initially.

If bulb \(B\) is removed from the circuit, the brightness of bulb \(C\) (A) remains the same. (B) goes up by a factor of four. (C) goes down by a factor of four. (D) goes down by a factor of \(4 / 9\). (E) goes up by a factor of \(9 / 4\).

For a rectangular block with \(L_1>L_2\), as drawn, can the resistance with the current through the length \(L_2\) be greater than the resistance with the current going through the length \(L_1\) ? (A) Always (B) Sometimes, if \(L_2\) is long enough (C) Sometimes, if \(A_2\) is large enough (D) Sometimes, if \(A_1\) is small enough (E) Never

If \(V=480 \mathrm{~V} ; C_1=10 \mu \mathrm{F} ; C_2=2 \mu \mathrm{F}\) and \(C_3=4 \mu \mathrm{F}\), rank the capacitors in terms of stored energy, from highest to lowest. (A) \(C_3, C_1=C_2\) (B) \(C_3, C_1, C_2\) (C) \(C_1, C_2, C_3\) (D) \(C_1=C_3, C_2\) (E) \(C_2, C_3, C_1\)
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