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Chapter 10: Problem 27

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) \(V_1=100 \mathrm{~V} ; V_2=400 \mathrm{~V}\)

Short Answer

Expert verified
Answer: (D) \(V_1=400V ; V_2=100V\)

Step by step solution

01

Find the equivalent capacitance for the series combination.

The equivalent capacitance for capacitors connected in series can be found using the formula: $$\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}$$ Substitute the given values of \(C_1\) and \(C_2\), and solve for \(C_{eq}\).
02

Calculate the current flowing through the circuit.

Using Ohm's law, we can find the current (\(I\)) flowing through the circuit, given by: $$I = \frac{V}{R_{eq}}$$ But for capacitors, we use the formula: $$I = C_{eq}(\frac{dV}{dt})$$ Since the problem statement does not describe the rate of change of voltage, we can assume it to be a constant DC voltage source. In this case, we can consider the current to be constant throughout the circuit.
03

Apply the voltage-divider rule to find the voltage drop across each capacitor.

The voltage-divider rule states that for a series circuit, the voltage across each component is proportional to its resistance (or impedance) and the total current. For capacitors, the impedance can be represented as \(\frac{1}{C}\). $$V_1 = \frac{V}{\frac{1}{C_1} + \frac{1}{C_2}} \cdot \frac{1}{C_1}$$ $$V_2 = \frac{V}{\frac{1}{C_1} + \frac{1}{C_2}} \cdot \frac{1}{C_2}$$
04

Calculate \(V_1\) and \(V_2\) using the given values of \(V\), \(C_1\), and \(C_2\).

Substitute the given values of \(V\), \(C_1\), and \(C_2\) into the equations from Step 3, and solve for \(V_1\) and \(V_2\). $$V_1 = \frac{500}{\frac{1}{2\mu F} + \frac{1}{8 \mu F}} \cdot \frac{1}{2\mu F} = 400 V$$ $$V_2 = \frac{500}{\frac{1}{2 \mu F} + \frac{1}{8\mu F}} \cdot \frac{1}{8\mu F} = 100 V$$
05

Compare the calculated values to the given options and find the correct answer.

Based on the calculated values, we find that option (D) \(V_1=400 \mathrm{~V} ; V_2=100 \mathrm{~V}\) is the correct answer.

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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 second resistor is halved to \(R\), the current dissipated power (A) stays the same. (B) goes up by a factor of two. (C) goes down by a factor of two. (D) is two-thirds what it was initially. (E) is three-halves what it was initially.

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) \(C_2, C_3, C_1\)

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 \(9 / 4\).

A rectangular copper block has a resistivity \(1.7 \times 10^{-8} \Omega \mathrm{m}\). \(L_1=15 \mathrm{~cm} ; L_2=3 \mathrm{~cm}\). The area of the faces are \(A_1=18 \mathrm{~cm}^2\) and \(A_2=90 \mathrm{~cm}^2\). A current is passed through the length \(L_1\). The resistance of the block is (A) \(1.4 \times 10^{-6} \Omega\) (B) \(1.4 \times 10^{-10} \Omega\) (C) \(5.7 \times 10^{-8} \Omega\) (D) \(5.7 \times 10^{-4} \Omega\) (E) \(7.1 \times 10^{-6} \Omega\)

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.
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