# Chapter 26 Direct-Current Circuits

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1 Chapter 26 Direct-Current Circuits 1 Resistors in Series and Parallel In this chapter we introduce the reduction of resistor networks into an equivalent resistor R eq. We also develop a method for analyzing more complicated circuit networks by using Kirchhoff s rules. In both these cases, the goal is to find the current between junctions in the circuits. Afterwards we will investigate the varying current that occurs in circuits containing both resistors and capacitors. Finally, we will take a brief look at power distribution systems. 1.1 Resistors in Series Figure 1: This figure shows three resistors in series. What is their equivalent resistance R eq? V ab = V ax + V xy + V yb = I (R 1 + R 2 + R 3 ) R eq = R 1 + R 2 + R 3 (Resistors in Series) (1) The equivalent resistance of a series combination equals the sum of the individual resistances. 1

2 Figure 2: This figure shows the different combinations one can have for resistors in series and resistors in parallel. 2

3 1.2 Resistors in Parallel I = I 1 + I 2 + I 3 = V ab ( 1 R R R 3 ) = V ab R eq Figure 3: This figure shows three resistors in parallel. What is their equivalent resistance R eq? 1 R eq = ( ) R 1 R 2 R 3 (Resistors in Parallel) (2) The reciprocal of the equivalent resistance of a parallel combination equals the sum of the reciprocals of the individual resistances. 3

4 Example: Calculate the power provided by the battery, and the power output of each incandescent light bulb (R = 2Ω) depending upon whether the light bulbs are powered as shown (a) in series or (b) in parallel. Figure 4: This figure shows two incandescent light bulbs, each with resistance R = 2Ω, powered by a single battery. In figure (a) the light bulbs are in series, while in figure (b) the light bulbs are in parallel. 4

5 2 Kirchhoff s Rules Many practical resistor networks cannot be reduced to simple series-parallel combinations. An example of this is shown in Fig. 5 where emf E 1 is charging a battery with a smaller emf E 2. Figure 5: This figure shows a more complicated circuit where two batteries are in a multi-loop circuit. This network cannot be reduced to a simple series-parallel combination of resistors. In this particular example emf E 1 is assumed to be charging emf E 2. To analyze these more complicated networks, we ll use the techniques developed by the German physicist Gustav Robert Kirchhoff ( ). First, we need to introduce two bits of terminology when analyzing these circuits. A junction in a circuit is a point where three of more conductors meet. A loop is any closed conducting path. 5

6 Kirchhoff s Junction Rule I = 0 (3) Figure 6: This figure shows three resistors in parallel. What is their equivalent resistance R eq? Kirchhoff s Loop Rule V = 0 (4) 2.1 Sign Conventions for the Loop Rule Figure 7: Use these sign conventions when you apply Kirchhoff s loop rule. The Travel direction going around the loop is not necessary the direction of the current. 6

7 Example: Figure 8: In this example we travel around the loop in the same direction as the assumed current, so all the IR terms are negative. Meanwhile, the 4V battery represents an emf-drop, and the 12V battery represents an emf-rise. 7

8 26.29: The V battery in Fig. E26.28 is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in in the figure. Find (a)!the current in each branch and (b) the potential difference V ab of point a relative to point b. Figure 9: This is Figure E26.28 for problem in University Physics 14 th edition. 8

9 3 Electrical Measuring Instruments 3.1 Ammeters 3.2 Voltmeters 3.3 Ohmmeters 3.4 The Potentiometer 9

10 4 R-C Circuits Up until now, we have assumed that all the emfs and resistances are constant (i.e., time independent) so that all the potentials, currents, and powers are also independent of time. However, by simply charging a capacitor in a circuit, we find that none of these physical quantities are independent of time. Figure 10: This figure shows (a) a capacitor completely uncharged before the switch is closed. Meanwhile figure (b) shows the capacitor in the process of being charged after the switch is closed. 4.1 Charging a Capacitor v ab = ir v bc = q C (voltage drops) 10

11 Using Krichhoff s loop rule we find: While the capacitor is being charged: E ir q C = 0 (5) i = E R q RC dq dt = 1 RC q(t) = Q 0 ( 1 e t/rc ) (q CE) (Charging the capacitor) (6) The instantaneous current can be found by taking a simple time-derivative. After the capacitor is charged the current i = 0. E R = Q 0 RC where Q 0 = CE the final charge Time Constant τ While investigating Eq. 6, we realize that the charge on the capacitor asymptotically approaches the final charge Q 0, but only as t. So, is there a rule of thumb (e.g., a time) that can be used to describe when a capacitor is close to being completely charged? The answer is, Yes, and it s called the time constant: τ = RC. The ruleof-thumb is that the capacitor is essentially charged when t = 5 τ = 5 RC. When t = 5RC = 5τ, the charge is q(5τ) = Q 0 (1 e 5 ) = 0.993Q 0, or 99.3% charged. Ex. 46: A 1.50-µF capacitor is charging through a 12.0-Ω resistor using a 12.0-V battery. (a) What will be the current when the capacitor has acquired 1/4 of its maximum charge? (b) Will it be 1/4 of the maximum current? 11

12 Figure 11: The second graph shows the instantaneous charge on the capacitor while it is charging up. This is described by Eq. 6. The first graph shows the instantaneous current i(t) = dq/dt. 4.2 Discharging a Capacitor Let s suppose we have a capacitor that is complete charged q = Q 0 and we discharge it. What is the charge on the capacitor as a function of time q(t). While the capacitor is being discharged, Kirchhoff s loop rule gives rise to the following equation: The solution to this equation is: i = dq dt = q RC q(t) = Q o e t/rc (Discharging the Capacitor) (7) Let s take a brief look at the power distribution in the circuit while charging the capacitor. Multiply Eq. 5 by i to obtain the instantaneous power P (t): ie = i 2 R + iq C (8) 12

13 Figure 12: Before the switch is closed at time t = 0, the capacitor is fully-charged with q = Q 0, and the current is zero as shown in figure (a). In figure (b), the switch is closed and the charge is released from the capacitor and defined by q(t). This results in a current that also varies as a function of time i(t). 13

14 5 Power Distribution Systems Figure 13: This schematic diagram shows part of a house wiring system. Only two branch circuits are shown; lamps and appliances may be plugged into the outlets. 14

15 5.1 Circuit Overloads and Short Circuits Figure 14: When a drill malfunctions when connected via a three-prong plug, a person touching it receives no shock, because electric charge flows through the ground wire (shown in green) to the third prong and into the local ground rather than into the person s body. 5.2 Household and Automotive Wiring 15

16 Homework Problems: Ex. 57: (a) Find the potential of point a with respect to point b in the figure. (b) If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 8.0V battery. Figure 15: This circuit is for Ex Ex. 78: A resistor with 900Ω is connected to the plates of a charged capacitor with capacitance 4.76µF. Just before the connection is made, the charge on the capacitor is 7.00mC. (a) What is the energy initially stored in the capacitor? (b) What is the electrical power dissipated in the resistor just after the connection is made? 16

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