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Chapter 24: Problem 44

The potential difference across two capacitors in series is \(120 . \mathrm{V}\). The capacitances are \(C_{1}=1.00 \cdot 10^{3} \mu \mathrm{F}\) and \(C_{2}=1.50 \cdot 10^{3} \mu \mathrm{F}\) a) What is the total capacitance of this pair of capacitors? b) What is the charge on each capacitor? c) What is the potential difference across each capacitor? d) What is the total energy stored by the capacitors?

Short Answer

Expert verified
Question: Calculate the total capacitance of the pair of capacitors, the charge on each capacitor, the potential difference across each capacitor, and the total energy stored by the capacitors, given a potential difference of 120V and capacitances \(C_1 = 1.00 \cdot 10^3 \mu F\) and \(C_2 = 1.50 \cdot 10^3 \mu F\). Answer: The total capacitance is 600 µF. The charge on each capacitor is 72 mC. The potential difference across \(C_1\) is 72V and across \(C_2\) is 48V. The total energy stored by the capacitors is 4.968 Joules.

Step by step solution

01

1. Calculate total capacitance

To determine the total capacitance of the two capacitors in series, we use the formula: $$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2}$$ Where \(C_1 = 1.00 \cdot 10^{3} \mu F\) and \(C_2 = 1.50 \cdot 10^{3} \mu F\). Solve for \(C_t\): $$C_t=\frac{1}{\frac{1}{1.00 \cdot 10^{3} \mu F}+\frac{1}{1.50 \cdot 10^{3} \mu F}}=600\mu F$$
02

2. Calculate charge on each capacitor

The charge is equal in capacitors connected in series. Therefore, we only need to find the charge on one capacitor: $$Q = C_t \cdot V$$ Where \(V = 120V\) and \(C_t = 600\mu F\). Solve for \(Q\): $$Q = 600\mu F \cdot 120 V = 72 mC$$ So, the charge on each capacitor is 72 mC.
03

3. Calculate potential difference across each capacitor

To find the potential difference across each capacitor, use the equation: $$V_i = \frac{Q}{C_i}\; \text{,where}\; i=1,2$$ For \(C_1 = 1.00 \cdot 10^{3} \mu F\) and \(C_2 = 1.50 \cdot 10^{3} \mu F\): $$V_1 = \frac{72 mC}{1.00 \cdot 10^{3} \mu F} = 72V$$ $$V_2 = \frac{72 mC}{1.50 \cdot 10^{3} \mu F} = 48V$$ The potential difference across \(C_1\) is 72V and across \(C_2\) is 48V.
04

4. Calculate total energy stored by the capacitors

To determine the total energy stored, we can sum up the energy stored in each capacitor: $$E = E_1 + E_2 = \frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2$$ With \(C_1 = 1.00 \cdot 10^{3} \mu F\), \(C_2 = 1.50 \cdot 10^{3} \mu F\), \(V_1 = 72V\) and \(V_2 = 48V\), we have: $$E = \frac{1}{2}(1.00 \cdot 10^{3} \mu F)(72V)^2 + \frac{1}{2}(1.50 \cdot 10^{3} \mu F)(48V)^2 = 3.24J + 1.728J = 4.968J$$ The total energy stored by the capacitors is 4.968 Joules.

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Most popular questions from this chapter

You have an electric device containing a \(10.0-\mu \mathrm{F}\) capacitor, but an application requires an \(18.0-\mu \mathrm{F}\) capacitor. What modification can you make to your device to increase its capacitance to \(18.0-\mu \mathrm{F} ?\)

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

A parallel plate capacitor of capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A close-fitting slab of dielectric, with dielectric constant \(\kappa\), is then inserted and fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.

A 4.00 -pF parallel plate capacitor has a potential difference of \(10.0 \mathrm{~V}\) across it. The plates are \(3.00 \mathrm{~mm}\) apart, and the space between them contains air. a) What is the charge on the capacitor? b) How much energy is stored in the capacitor? c) What is the area of the plates? d) What would the capacitance of this capacitor be if the space between the plates were filled with polystyrene?

A parallel plate capacitor is connected to a battery. As the plates are moved farther apart, what happens to each of the following? a) the potential difference across the plates b) the charge on the plates c) the electric field between the plates
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