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Two capacitors are connected in parallel across the terminals of a battery. One has a capacitance of \(2.0 \mu \mathrm{F}\) and the other a capacitance of \(4.0 \mu F .\) These two capacitors together store \(5.4 \times 10^{-5} \mathrm{C}\) of charge. What is the voltage of the battery?

Short Answer

Expert verified
The voltage of the battery is 9.0 volts.

Step by step solution

01

Understand Capacitors in Parallel

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. Denoting the capacitance of the first and second capacitors as \( C_1 \) and \( C_2 \) respectively, we have \( C_1 = 2.0 \mu F \) and \( C_2 = 4.0 \mu F \). The total capacitance \( C_{total} \) is given by \( C_{total} = C_1 + C_2 \).
02

Calculate Total Capacitance

Calculate the total capacitance of the capacitors:\[C_{total} = C_1 + C_2 = 2.0 \mu F + 4.0 \mu F = 6.0 \mu F\]
03

Understand the Relationship Between Charge, Voltage, and Capacitance

The relationship between charge \( Q \), voltage \( V \), and capacitance \( C \) is given by \( Q = C \times V \). In this problem, we are asked to find the voltage \( V \) across the capacitors, using the total charge \( Q = 5.4 \times 10^{-5} C \).
04

Solve for Voltage

Rearrange the formula \( Q = C \times V \) to solve for \( V \):\[V = \frac{Q}{C_{total}}\]Substitute the known values:\[V = \frac{5.4 \times 10^{-5} C}{6.0 \times 10^{-6} F} = 9.0 V\]
05

Conclusion

The voltage of the battery is 9.0 volts. This is the potential difference across the terminals of the capacitors in parallel.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Total Capacitance
When dealing with capacitors connected in parallel, it is important to understand how the total capacitance of the circuit is calculated. In a parallel configuration, each capacitor has the same voltage across it. Therefore, the overall capacitance is simply the sum of the individual capacitances. This is because adding capacitors in parallel increases the surface area for storing charge, and thus a larger total capacitance is obtained.

For example, if you have two capacitors in parallel, with capacitances of \(2.0 \mu F\) and \(4.0 \mu F\), the total capacitance \(C_{total}\) can be calculated as follows:

  • Total capacitance formula: \(C_{total} = C_1 + C_2\)
  • Calculation: \(C_{total} = 2.0 \mu F + 4.0 \mu F = 6.0 \mu F\)

This concept is very useful in designing circuits where a certain capacitance is needed, as you can easily achieve a desired total capacitance by combining capacitors in parallel.
Charge-Voltage-Capacitance Relationship
The fundamental relationship between charge, voltage, and capacitance is pivotal in understanding how capacitors function in a circuit. The equation \(Q = C \times V\) describes this relationship, where \(Q\) is the stored charge, \(C\) is the capacitance, and \(V\) is the voltage across the capacitor. This means that the charge stored in a capacitor is directly proportional to the voltage applied and its capacitance.

To find the voltage when you have a certain charge and capacitance, rearrange the formula as follows:
  • Formula for voltage: \(V = \frac{Q}{C}\)

In practice, if a parallel circuit of two capacitors stores \(5.4 \times 10^{-5} C\) of charge with a total capacitance of \(6.0 \mu F\), you can find the voltage using:
  • \(V = \frac{5.4 \times 10^{-5} C}{6.0 \times 10^{-6} F} = 9.0 V\)

This calculated voltage corresponds to the potential difference across the capacitors.
Series and Parallel Circuits
In the study of electrical circuits, knowing how capacitors behave in series versus parallel arrangements is critical. Each arrangement has distinct principles and calculations.

Parallel Circuits:
  • The voltage across each capacitor is the same.
  • Total capacitance is the sum of all individual capacitances \(C_{total} = C_1 + C_2 + ... + C_n\)
  • Used when you need a higher capacitance in a circuit.

Series Circuits:
  • The same charge flows through each capacitor.
  • Total capacitance is found by the inverse sum of the reciprocals: \(\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}\)
  • Used when decreasing the effective capacitance is necessary.

Understanding these principles ensures that you can configure capacitance in a circuit to meet specific requirements effectively, whether it's to increase energy storage or adjust for voltage levels.

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Most popular questions from this chapter

There are approximately 110 million households that use TVs in the United States. Each TV uses, on average, \(75 \mathrm{W}\) of power and is turned on for 6.0 hours a day. If electrical energy costs \(\$ 0.12\) per \(\mathrm{kWh}\), how much money is spent every day in keeping 110 million TVs turned on?

The total current delivered to a number of devices connected in parallel is the sum of the individual currents in each device. Circuit breakers are resettable automatic switches that protect against a dangerously large total current by "opening" to stop the current at a specified safe value. A \(1650-\mathrm{W}\) toaster, a \(1090-\mathrm{W}\) iron, and a \(1250-\mathrm{W}\) microwave oven are turned on in a kitchen. As the drawing shows, they are all connected through a \(20-\mathrm{A}\) circuit breaker (which has negligible resistance) to an ac voltage of \(120 \mathrm{V}\). (a) Find the equivalent resistance of the three devices. (b) Obtain the total current delivered by the source and determine whether the breaker will "open" to prevent an accident.

A \(75.0-\Omega\) and a \(45.0-\Omega\) resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is \(0.294 \mathrm{A}\). When the \(45.0-\Omega\) resistor is disconnected, the current from the battery drops to 0.116 A. Determine (a) the emf and (b) the internal resistance of the battery.

A fax machine uses 0.110 A of current in its normal mode of operation, but only \(0.067 \mathrm{A}\) in the standby mode. The machine uses a potential difference of \(120 \mathrm{V}\). In one minute (a) how much more charge passes through the machine in the normal mode than in the standby mode, and (b) how much more energy is used?

Three capacitors are connected in series. The equivalent capacitance of this combination is \(3.00 \mu\) F. Two of the individual capacitances are \(6.00 \mu \mathrm{F}\) and \(9.00 \mu \mathrm{F}\). What is the third capacitance (in \(\mu \mathrm{F}\) )?

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