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Chapter 10: Problem 29

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) \(C_2, C_3, C_1\)

Short Answer

Expert verified
Question: Rank the capacitors from highest to lowest stored energy based on the values \(C_1=10 \mu \mathrm{F}, C_2=2 \mu \mathrm{F}\) and \(C_3=4 \mu \mathrm{F}\) and voltage \(V=480 \mathrm{V}\) across each capacitor. Answer: (C) \(C_1, C_3, C_2\)

Step by step solution

01

Calculating energy stored in Capacitor C1

Calculate the energy stored in C1 using the formula: \(E = \frac{1}{2} CV^2\). In this case, C1 is \(10 \mu \mathrm{F}\) and V is \(480 \mathrm{V}\). So, \(E_{C1} = \frac{1}{2} (10 \mu \mathrm{F}) (480 \mathrm{V})^2\)
02

Calculating energy stored in Capacitor C2

Calculate the energy stored in C2 using the formula: \(E = \frac{1}{2} CV^2\). In this case, C2 is \(2 \mu \mathrm{F}\) and V is \(480 \mathrm{V}\). So, \(E_{C2} = \frac{1}{2} (2 \mu \mathrm{F}) (480 \mathrm{V})^2\)
03

Calculating energy stored in Capacitor C3

Calculate the energy stored in C3 using the formula: \(E = \frac{1}{2} CV^2\). In this case, C3 is \(4 \mu \mathrm{F}\) and V is \(480 \mathrm{V}\). So, \(E_{C3} = \frac{1}{2} (4 \mu \mathrm{F}) (480 \mathrm{V})^2\)
04

Comparing energies and ranking capacitors

Now that we have calculated the energy stored in each capacitor, we can rank them from highest to lowest energy: - \(E_{C1} = \frac{1}{2} (10 \mu \mathrm{F}) (480 \mathrm{V})^2 = 1.152 \times 10^{6} \mu \mathrm{J}\) - \(E_{C2} = \frac{1}{2} (2 \mu \mathrm{F}) (480 \mathrm{V})^2 = 230400 \mu \mathrm{J}\) - \(E_{C3} = \frac{1}{2} (4 \mu \mathrm{F}) (480 \mathrm{V})^2 = 460800 \mu \mathrm{J}\) Based on the calculated energies, we can rank the capacitors as follows: \(C_1, C_3, C_2\) Thus, the correct answer is (C) \(C_1, C_2, C_3\).

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Most popular questions from this chapter

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.

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.

Two capacitors are connected in series to a voltage source, as shown below. If \(V=500 ; C_1=2 \mu \mathrm{F}\) and \(C_2=8 \mu \mathrm{F}\), what is the voltage drop, \(V_1\) and \(V_2\), across each? (A) \(V_1=250 \mathrm{~V} ; V_2=250 \mathrm{~V}\) (B) \(V_1=20 \mathrm{~V} ; V_2=80 \mathrm{~V}\) (C) \(V_1=200 \mathrm{~V} ; V_2=300 \mathrm{~V}\) (D) \(V_1=400 \mathrm{~V} ; V_2=100 \mathrm{~V}\) (E) \(V_1=100 \mathrm{~V} ; V_2=400 \mathrm{~V}\)

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

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
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