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Chapter 21: Electric Currents and Direct-Current Circuits Chapter Review |
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Chapter Review
21-6 Circuits Containing Capacitors
Capacitors are commonly used in circuits and can be combined in series and parallel producing an overall equivalent capacitance Ceq that depends on the type of connection. As with resistors,
capacitors in parallel are connected across the same potential difference.
Thinking of a parallel-plate capacitor
, connecting capacitors in parallel is like making the plates of one capacitor wider. The result is that for N caoacitors connected in parallel the equivalent capacitance is
.
Therefore, in parallel, the equivalent capacitance is larger than any of the individual capacitances that contribute to it.
Just as resistors in series have the same current flowing through them, capacitors in series have the same current flow onto them and thus each capacitor stores the same amount of charge Q. Therefore,
capacitors in series store equal amounts of charge.
Connecting capacitors in series has the same effect of increasing the distance between the plates of one capacitor. The result for N capacitors connected in series is that the equivalent capacitance can be determined by
.
Once 1/Ceq is determined we can invert it to determine Ceq in farads. Therefore, in series, the equivalent capacitance is smaller than the smallest of the individual capacitances that contribute to it.
Physlet Illustration: Capacitors in Parallel | |
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| In this simulation, a 10 microFarad, 15 microFarad, and 25 microFarad capacitor are connected in parallel. Move the mouse over the capacitors to measure their charge. How is the sum of these charges related to the charge on the equivalent capacitor? | |
Hints:
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Example 21.5 Capacitors: Determine both the charge on and the potential difference across each capacitor in the circuit below given that C1 = 12 mF, C2 = 5.0 mF, C3 = 6.5 mF, and E = 12 V.
Picture the Problem The picture shows the circuit in question.
Strategy We can use the known facts about capacitors in series and parallel to obtain the required information.
Solution
| 1. The equivalent capacitance of C1 and C2 is: | C12 = C1 + C2 = 12 mF + 5.0 mF = 17.0 mF |
| 2. The overall equivalent capacitance is: | ![]() |
| 3. The total charge stored is: | Qtot = CeqV = (4.702 mF)(12 V) = 56.43 mC |
| 4. Since the same charge is stored on capacitors in series the charge stored on C3 is: | Q3 = Qtot = 56 mC |
| 5. The potential difference across C3 is: | ![]() |
| 6. By the loop rule, the potential difference across the C1 and C2 combination is: | V12 = E - V3 = 12 V - 8.681 V = 3.319 V |
| 7. Since capacitors in parallel are across the same potential difference: | V1 = V2 = 3.3 V |
| 8. The charge stored on C1 is: | Q1 = C1V1 = (12 mF)(3.319 V) = 40 mC |
| 9. The charge stored on C2 is: | Q2 = C2V2 = (5.0 mF)(3.319 V) = 17 mC |
Insight Make sure you understand the results of steps 4 and 7; these steps are the keys to the solution.
Physlet Illustration: Capacitors in Series | |
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| In this simulation, a 10 microFarad, 15 microFarad, and 30 microFarad capacitor are connected in series. Move the mouse over the capacitors to measure their charge. How are these related to the charge on the equivalent capacitor? | |
Hints:
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Practice Quiz
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21-7 RC Circuits
For our purposes an RC circuit is one that contains resistance and capacitance in series. In these circuits the charge on the capacitor builds up (and dies off) slowly in comparison to what happens in circuits containing only resistance and no capacitance. For an RC circuit in which the capacitor is initially uncharged, the charge builds up on the capacitor according to the equation
where E is the source emf. The quantity t is given by t = RC and is called the time constant of the circuit. The time constant provides a characteristic amount of time for the charge to build because after an amount of time t = t the capacitor will be most of the way (63.2 %) to its maximum charge of qmax = CE. In the present case of an initially uncharged capacitor, the current starts out at its maximum value and falls off exponentially
If the capacitor is initially charged with a charge Q, has a potential difference V across it, and is allowed to discharge, both the charge and current fall off exponentially
Example 21.6 An RC Circuit: Determine the time constant of the RC circuit shown below.
Picture the Problem The picture shows the circuit in question.
Strategy We need to determine the correct values of Req and Ceq and then use them to get t.
Solution
| 1. The equivalent resistance is: | Req = R1 + R2 = 2.0 W + 3.0 W = 5.0 W |
| 2. The equivalent capacitance is: | Ceq = C1 + C2 = 4.0 mF + 1.0 mF = 5.0 mF |
| 3. The time constant is: | t = ReqCeq = (5.0 W)(5.0 mF) = 25 ms |
Insight Here, despite the fact that there are two resistors and two capacitors, we still have resistance and capacitance in series.
Physlet Illustration: RC Circuit |
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| In this simulation, a resistor and a capacitor are connected in series to a 12-Volt battery. When the switch closes, the capacitor begins to charge. A voltmeter is connected across the capacitor, and a graph of the voltage across the capacitor vs. time is also shown. Vary the values of the resistor (1 W < R < 10 W) and capacitor (10 mf < C < 50 mf). How do these affect the charging of the capacitor? |
Hints:
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Practice Quiz
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*21-8 Ammeters and Voltmeters
An ammeter is a device specially made for measuring currents in a circuit. An ammeters should be connected in series with the device whose current is sought. Ideally, an ammeter should have zero resistance. In practice, the resistance of an ammeter should be much less than the resistances of other devices in the circuit.
A voltmeter is specially designed to measure potential differences across devices in a circuit. A voltmeter should be connected in parallel with the device across which the potential difference is being sought. Ideally, a voltmeter should have infinite resistance. In practice, the resistance of a voltmeter should be much greater than the resistances of other devices in the circuit.
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