Chapter 28: 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.

Sources of “Electromotive Force” or EMF. EMF & Terminal Voltage An electric circuit needs a battery or a generator to produce current – these are called Sources of “Electromotive Force” or EMF. It is important to remember that, despite its misleading name, EMF is a VOLTAGE source! It is NOT a FORCE!! A battery is a nearly constant voltage source, but it does have a small internal resistance r, which reduces the actual voltage from the ideal EMF:

The internal battery resistance r behaves as if it were in series with the EMF. Figure 26-1. Diagram for an electric cell or battery.

Electromotive Force The electromotive force (emf), e, of a battery is the maximum possible voltage that the battery can provide between its terminals. The emf supplies energy, it does not apply a force. The battery will normally be the source of Energy in the circuit. The positive terminal of the battery is at a higher potential than the negative terminal. We assume that the wires to have no resistance.

Example: Battery with Internal Resistance A resistor R = 65.0-Ω is connected to the terminals of a battery with emf E =12.0 V & internal resistance r = 0.5 Ω. Calculate (a) the current I in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R & in the battery’s internal resistance r. Solution: a. The total resistance is 65.5 Ω, so the current is 0.183 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.

Example: Battery with Internal Resistance A resistor R = 65.0-Ω is connected to the terminals of a battery with emf E =12.0 V & internal resistance r = 0.5 Ω. Calculate (a) the current I in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R & in the battery’s internal resistance r. Answers: (a) I = 0.183 A. (b) Vab = 11.9 V. (c) PR = 2.18W, Pr = 0.02 W Solution: a. The total resistance is 65.5 Ω, so the current is 0.183 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.

Resistors in Series & 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.

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:

This can be used to obtain the Equivalent Resistance Req (Req  The single resistance that gives the same current in the circuit for the same voltage drop.) V  IReq Figure 26-3c. Equivalent single resistance that draws the same current: Req = R1 + R2 + R3 .

A PARALLEL CONNECTION splits the current; the voltage across each resistor is the same: Figure 26-4. (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

The total current I is the sum of currents across each resistor: , This gives the reciprocal of the equivalent resistance Req I  (V/Req)

An analogy using water may be helpful in visualizing parallel circuits 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 26-4. Water pipes in parallel—analogy to electric currents in parallel.

Conceptual Example: Series or parallel? (a) The light bulbs 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.

Conceptual Example: 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).

Circuit with series and parallel resistors. Example: Circuit with series and parallel resistors. Calculate the current that is drawn from the battery shown. 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.

Example: Current in one branch. Calculate the current I1 through the 500-Ω resistor. This is the same circuit as in the previous problem.) The total current I in the circuit was found to be I = 17 mA. Solution: The total current is 17 mA, so the voltage drop across the 400-Ω resistor is V = IR = 7.0 V. This means that the voltage drop across the 500-Ω resistor is 5.0 V, and the current through it is I = V/R = 10 mA.

Bulb Brightness in a Circuit. Use a minimum of mathematics Conceptual Example Bulb Brightness in a Circuit. The circuit shown has 3 identical light bulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A & B compare with that of C? (b) What happens when switch S is opened? 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. Use a minimum of mathematics to obtain your answer!!!

Example: 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.

Example: Analyzing a Circuit A 9.0-V battery with internal resistance r = 0.50 Ω is connected in the circuit shown. (a) How much current is drawn from the battery? (b) What is the battery terminal voltage? (c) What is the current in the 6.0-Ω resistor? 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.