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Chapter 23: Problem 29

You have a \(1.0-\mu \mathrm{F}\) and a \(2.0-\mu \mathrm{F}\) capacitor. What capacitances can you get by connecting them in series or in parallel?

Short Answer

Expert verified
The total capacitance when the capacitors are connected in series is \(2/3 \mu F\) and when the capacitors are connected in parallel, the total capacitance is \(3.0 \mu F\).

Step by step solution

01

Capacitance in Series

Calculate the total capacitance \(C_{total}\) when the capacitors are connected in series. Using the formula for capacitors in series, \(1/{C_{total}} = 1/{C_1} + 1/{C_2}\). So, \(1/{C_{total}} = 1/1.0 \mu F + 1/2.0 \mu F = 1.5\). Hence, \(C_{total} = 2/3 \mu F\).
02

Capacitance in Parallel

Next, calculate the total capacitance \(C_{total}\) when the capacitors are connected in parallel. Using the formula for capacitors in parallel, \(C_{total} = {C_1} + {C_2}\). So, \(C_{total} = 1.0 \mu F + 2.0 \mu F = 3.0 \mu F\).

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Most popular questions from this chapter

A student argues that the total energy associated with the electric field of a charged sphere must be infinite because its field extends throughout an infinite volume. Critique this argument.

Which can store more energy: a \(1.0-\mu \mathrm{F}\) capacitor rated at \(250 \mathrm{V}\) or a 470 -pF capacitor rated at \(3 \mathrm{kV} ?\)

A solid sphere contains a uniform volume charge density. What fraction of the total electrostatic energy of this configuration is contained within the sphere?

Charge is spread over the surface of a balloon, which is then allowed to expand. What happens to the energy of the electric field?

An unknown capacitor \(C\) is connected in series with a \(3.0-\mu \mathrm{F}\) capacitor; this pair is placed in parallel with a 1.0 - \(\mu\) F capacitor, and the entire combination is put in series with a \(2.0-\mu \mathrm{F}\) capacitor. (a) Make a circuit diagram of this network. (b) When a potential difference of \(100 \mathrm{V}\) is applied across the open ends of the network, the total energy stored in all the capacitors is \(5.8 \mathrm{mJ} .\) Find \(C\).
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