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DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram below shows a possible amplifier circuit for each stereo channel. Although the large triangle is an amplifier chip containing transistors (discussed in Chapter 40), the other circuit elements are ones we have met, resistors and capacitors, and we discuss them in circuits in this Chapter. We also discuss voltmeters and ammeters, and how they are built and used to make measurements.

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**Homework Chap. 28 Exer. 5, 6, 7, 8, Exer. 9, 10, 11, 12, 17, 18,**

Prob. 2.

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**EMF and Terminal Voltage**

Resistors in Series and in Parallel Kirchhoff’s Rules Series and Parallel EMFs; Battery Charging Circuits Containing Resistor and Capacitor (RC Circuits) Electric Hazards Ammeters and Voltmeters

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**EMF and Terminal Voltage**

Electric circuit needs battery or generator to produce current – these are called sources of emf (Electromotive force). Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf: Terminal Voltage emf

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**EMF and Terminal Voltage**

This resistance behaves as though it were in series with the emf. Figure Diagram for an electric cell or battery.

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**EMF and Terminal Voltage**

Battery with internal resistance. A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r. Solution: a. The total resistance is 65.5 Ω, so the current is A. b. The voltage is the emf less the voltage drop across the internal resistance: V = 11.9 V. c. The power is I2R; in the resistor it is 2.18 W and in the internal resistance it is 0.02 W.

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**Resistors in Series and in Parallel**

A series connection has a single path from the battery, through each circuit element in turn, then back to the battery. Figure 26-3a. Resistances connected in series.

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**Resistors in Series and in Parallel**

The current through each resistor is the same The voltage depends on the resistance. The sum of the voltage drops across the resistors equals the battery voltage:

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**Resistors in Series and in Parallel**

A parallel connection splits the current; the voltage across each resistor is the same: Figure (a) Resistances connected in parallel. (b) The resistances could be lightbulbs. (c) The equivalent circuit with Req obtained from Eq. 26–4: 1/Req = 1/R1 + 1/R2 + 1/R3

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**Resistors in Series and in Parallel**

The voltage across each resistor is the same: The total current is the sum of the currents across each resistor: ,

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**Resistors in Series and in Parallel**

This gives the reciprocal of the equivalent resistance:

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**Resistors in Series and in Parallel**

An analogy using water may be helpful in visualizing parallel circuits. The water (current) splits into two streams; each falls the same height, and the total current is the sum of the two currents. With two pipes open, the resistance to water flow is half what it is with one pipe open. Figure Water pipes in parallel—analogy to electric currents in parallel.

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**Resistors in Series and in Parallel**

Series or parallel? (a) The lightbulbs in the figure are identical. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current. Solution: a. In the parallel configuration, the equivalent resistance is less, so the current is higher and the lights will be brighter. b. They are wired in parallel, so that if one light burns out the other one still stays on.

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**Resistors in Series and in Parallel**

An illuminating surprise. A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb are connected in two different ways as shown. In each case, which bulb glows more brightly? Ignore change of filament resistance with current (and temperature). Solution: a. Each bulb sees the full 120V drop, as they are designed to do, so the 100-W bulb is brighter. b. P = V2/R, so at constant voltage the bulb dissipating more power will have lower resistance. In series, then, the 60-W bulb – whose resistance is higher – will be brighter. (More of the voltage will drop across it than across the 100-W bulb).

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**Resistors in Series and in Parallel**

Circuit with series and parallel resistors. How much current is drawn from the battery shown? What is the current through each of the resistor? Solution: First we find the equivalent resistance of the two resistors in parallel, then the series combination of that with the third resistance. For the two parallel resistors, R = 290 Ω (remember that the usual equation gives the INVERSE of the resistance – students are often confused by this). The series combination is then 690 Ω, so the current in the battery is V/R = 17 mA.

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**Resistors in Series and in Parallel**

Bulb brightness in a circuit. The circuit shown has three identical lightbulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A and B compare with that of bulb C? (b) What happens when switch S is opened? Use a minimum of mathematics in your answers. Solution: a. When S is closed, the bulbs in parallel have an equivalent resistance equal to half that of the series bulb. Therefore, the voltage drop across them is smaller. In addition, the current splits between the two of them. Bulbs A and B will be equally bright, but much dimmer than C. b. With switch S open, no current flows through A, so it is dark. B and C are now equally bright, and each has half the voltage across it, so C is somewhat dimmer than it was with the switch closed, and B is brighter.

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**Resistors in Series and in Parallel**

A two-speed fan. One way a multiple-speed ventilation fan for a car can be designed is to put resistors in series with the fan motor. The resistors reduce the current through the motor and make it run more slowly. Suppose the current in the motor is 5.0 A when it is connected directly across a 12-V battery. (a) What series resistor should be used to reduce the current to 2.0 A for low-speed operation? (b) What power rating should the resistor have? Solution: a. The resistance of the motor is V/I = 2.4 Ω. For the current to be 2.0 A, the voltage across the motor must be 4.8 V; the other 7.2 V must appear across the resistor, whose resistance is therefore 7.2V/2.0A = 3.6 Ω. b. The power is IV = 14.4 W; a 20-W rating would be reasonably safe.

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**Resistors in Series and in Parallel**

Analyzing a circuit. A 9.0-V battery whose internal resistance r is 0.50 Ω is connected in the circuit shown. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? (c) What is the current in the 6.0-Ω resistor? a b c d Solution: a. First, find the equivalent resistance. The 8-Ω and 4-Ω resistors in parallel have an equivalent resistance of 2.7 Ω; this is in series with the 6.0-Ω resistor, giving an equivalent of 8.7 Ω. This is in parallel with the 10.0-Ω resistor, giving an equivalent of 4.8 Ω; finally, this is in series with the 5.0-Ω resistor and the internal resistance of the battery, for an overall resistance of 10.3 Ω. The current is 9.0 V/10.3 Ω = 0.87 A. b. The terminal voltage of the battery is the emf less Ir = 8.6 V. c. The current across the 6.0-Ω resistor and the 2.7-Ω equivalent resistance is the same; the potential drop across the 10.0-Ω resistor and the 8.3-Ω equivalent resistor is also the same. The sum of the potential drop across the 8.3-Ω equivalent resistor plus the drops across the 5.0-Ω resistor and the internal resistance equals the emf of the battery; the current through the 8.3-Ω equivalent resistor is then 0.48 A.

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Kirchhoff’s Rules Some circuits cannot be broken down into series and parallel connections. For these circuits we use Kirchhoff’s rules. Figure Currents can be calculated using Kirchhoff’s rules.

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Kirchhoff’s Rules Junction rule: The sum of currents entering a junction equals the sum of the currents leaving it.

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Kirchhoff’s Rules Loop rule: The sum of the changes in potential around a closed loop is zero. Figure Changes in potential around the circuit in (a) are plotted in (b).

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Kirchhoff’s Rules Junction rule: Loop rule:

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**Kirchhoff’s Rules Label each current, including its direction.**

Identify unknowns. Apply junction and loop rules; you will need as many independent equations as there are unknowns. Solve the equations, being careful with signs. If the solution for a current is negative, that current is in the opposite direction from the one you have chosen.

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**Kirchhoff’s Rules Using Kirchhoff’s rules.**

Calculate the currents I1, I2, and I3 in the three branches of the circuit in the figure. Solution: You will have two loop rules and one junction rule (there are two junctions but they both give the same rule, and only 2 of the 3 possible loop equations are independent). Algebraic manipulation will give I1 = A, I2 = 2.6 A, and I3 = 1.7 A.

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**Homework Chap. 28 Chap. 29 Exer. 35, 36, 37, 38, Exer. 45, 46, 47, 48,**

Prob. 7. Chap. 29 Exer. 3, 4, 5, 6, 7, 8, Exer. 13, 14, 15, 16, Exer. 20, 21, 24,

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**Calculate the equivalent**

resistance:

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**Series and Parallel EMFs; Battery Charging**

EMFs in series in the same direction: total voltage is the sum of the separate voltages.

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**Series and Parallel EMFs**

EMFs in series, opposite direction: total voltage is the difference, but the lower-voltage battery is charged.

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**Series and Parallel EMFs**

EMFs in parallel only make sense if the voltages are the same; this arrangement can produce more current than a single emf.

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**Battery Charging Jump starting a car.**

A good car battery is being used to jump start a car with a weak battery. The good battery has an emf of 12.5 V and internal resistance Ω. Suppose the weak battery has an emf of 10.1 V and internal resistance 0.10 Ω. Each copper jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown. Assume the starter motor can be represented as a resistor Rs = 0.15 Ω. Determine the current through the starter motor (a) if only the weak battery is connected to it, and (b) if the good battery is also connected. Solution: a. The weak battery is connected to two resistances in series; I = 40 A. b. First, find the resistance of the jumper cables; it is Ω. Now this is a problem similar to Example 26-9; you will need three equations to find the three (unknown) currents. (Technically you only have to find the current through the starter motor, but you still need three equations). The current works out to be 71 A.

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RC Circuits When the switch is closed, the capacitor will begin to charge. As it does, the voltage across it increases, and the current through the resistor decreases. Figure After the switch S closes in the RC circuit shown in (a), the voltage across the capacitor increases with time as shown in (b), and the current through the resistor decreases with time as shown in (c).

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**Circuits Containing Resistor and Capacitor (RC Circuits)**

If an isolated charged capacitor is connected across a resistor, it discharges: the voltage across it decreases. Figure For the RC circuit shown in (a), the voltage VC across the capacitor decreases with time, as shown in (b), after the switch S is closed at t = 0. The charge on the capacitor follows the same curve since VC α Q.

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**Circuits Containing Resistor and Capacitor (RC Circuits)**

To find the voltage as a function of time, we write the equation for the voltage changes around the loop:

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**RC Circuits For discharging process, I= -dQ/dt:**

At t=0,Q= Q0; therefore k=lnQ0 :

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**RC Circuits For charging process, I= +dQ/dt:**

At t=0,Q= 0; therefore k=ln(CE) :

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RC Circuits The quantity RC that appears in the exponent is called the time constant of the circuit: Another time is half-life, T1/2:

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**RC Circuits Discharging RC circuit.**

In the RC circuit shown, the battery has fully charged the capacitor, so Q0 = CE. Then at t = 0 the switch is thrown from position a to b. The battery emf is 20.0 V, and the capacitance C = 1.02 μF. The current I is observed to decrease to 0.50 of its initial value in 40 μs. (a) What is the value of Q, the charge on the capacitor, at t = 0? (b) What is the value of R? (c) What is Q at t = 60 μs? Solution: a. At t = 0, Q = CE = 20.4 μC. b. At t = 40 μs, I = 0.5 I0. Substituting in the exponential decay equation gives R = 57 Ω. c. Q = 7.3 μC.

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**RC Circuits RC circuit, with emf.**

The capacitance in the circuit shown is C = 0.30 μF, the total resistance is 20 kΩ, and the battery emf is 12 V. Determine (a) the time constant, (b) the maximum charge the capacitor could acquire, (c) the time it takes for the charge to reach 99% of this value, (d) the current I when the charge Q is half its maximum value, (e) the maximum current, and (f) the charge Q when the current I is 0.20 its maximum value. Solution: a. The time constant is RC = 6.0 x 10-3 s. b. The maximum charge is the emf x C = 3.6 μC. c. Set Q(t) = 0.99 Qmax and solve for t: t = 28 ms. d. When Q = 1.8 μC, I = 300 μA. e. The maximum current is the emf/R = 600 μA. f. When I = 120 μA, Q = 2.9 μC.

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**Circuits Containing Resistor and Capacitor (RC Circuits)**

Bulb in RC circuit. In the circuit shown, the capacitor is originally uncharged. Describe the behavior of the lightbulb from the instant switch S is closed until a long time later. Solution: When the switch is closed, the current is large and the bulb is bright. As the capacitor charges, the bulb dims; once the capacitor is fully charged the bulb is dark.

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**Ammeters and Voltmeters**

An ammeter measures current; a voltmeter measures voltage. Both are based on galvanometers, unless they are digital. The current in a circuit passes through the ammeter; the ammeter should have low resistance so as not to affect the current. Figure An ammeter is a galvanometer in parallel with a (shunt) resistor with low resistance, Rsh.

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**Ammeters and Voltmeters**

Ammeter design. Design an ammeter to read 1.0 A at full scale using a galvanometer with a full-scale sensitivity of 50 μA and a resistance r = 30 Ω. Check if the scale is linear. Solution: When the current is 1 A, 50 μA should flow through the ammeter. The shunt resistor is in parallel with the galvanometer; using Ohm’s law we find that the shunt resistance should be 1.5 x 10-3 Ω. The scale will be linear as long as the current does not affect the resistance.

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**Ammeters and Voltmeters**

A voltmeter should not affect the voltage across the circuit element it is measuring; therefore its resistance should be very large. Figure A voltmeter is a galvanometer in series with a resistor with high resistance, Rser.

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**Ammeters and Voltmeters**

Voltmeter design. Using a galvanometer with internal resistance 30 Ω and full-scale current sensitivity of 50 μA, design a voltmeter that reads from 0 to 15 V. Is the scale linear? Solution: When the voltage across the galvanometer is 15V, there should be 50 μA of current flowing through it. Using Ohm’s law, we find the series resistance should be 300 kΩ. The scale will again be linear.

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**Ammeters and Voltmeters**

An ohmmeter measures resistance; it requires a battery to provide a current. Figure An ohmmeter.

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**Ammeters and Voltmeters**

Summary: An ammeter must be in series with the current it is to measure; a voltmeter must be in parallel with the voltage it is to measure.

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**Ammeters and Voltmeters**

Voltage reading vs. true voltage. Suppose you are testing an electronic circuit which has two resistors, R1 and R2, each 15 kΩ, connected in series as shown in part (a) of the figure. The battery maintains 8.0 V across them and has negligible internal resistance. A voltmeter whose sensitivity is 10,000 Ω/V is put on the 5.0-V scale. What voltage does the meter read when connected across R1, part (b) of the figure, and what error is caused by the finite resistance of the meter? Solution: On the 5-V scale, the meter has a resistance of 50,000 Ω. The equivalent resistance of the meter and R1 in parallel is 11.5 kΩ. The total resistance of the circuit is 26.5 kΩ, so the current is 0.30 mA; the voltage across R1 (and therefore also across the voltmeter) is 3.5 V. Without the voltmeter, the voltage across R1 is half the battery voltage, or 4.0 V; this is more than a 10% error.

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Summary A source of emf transforms energy from some other form to electrical energy. A battery is a source of emf in parallel with an internal resistance. Resistors in series:

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**Summary Resistors in parallel: Kirchhoff’s rules:**

Sum of currents entering a junction equals sum of currents leaving it. Total potential difference around closed loop is zero.

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**Summary RC circuit has a characteristic time constant:**

Ammeter: measures current. Voltmeter: measures voltage.

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