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27: Chapter Circuits Problem 1 A wire of resistance 5.0 is connected to a battery whose emf E is 2.0 V and whose internal resistance is 1.0 . In 2.0 min, how much energy is (a) transferred from chemical to electrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery? Register to View Answer80 J; (b) 66.7 J; (c) 13.3 J Solution (a) The energy transferred from chemical to electrical in the battery is 2t U = Pt = it = Rtot Where the total resistance for the wire and internal resistance in series is Rtot = 5.0 + 1.0 = 6.0 Therefore ( 2.0V )2 2 min = 80J U= 6 (b) The thermal energy in the 5.0- wire is 2.0V U = Pt = i Rwiret = Rwiret = 6.0 5.0 2.0 min = 66.7J Rtot 2 2 2 (c) The "missing" 13.3 J = 80 J 66.7 J is dissipated as thermal energy because of the internal resistance. Problem 3 A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced? Answer 10.8 kJ Solution The energy transferred is the product of the power and the time and the power is the product of the current and the potential difference. U = iEt = (5.0 A)*(6.0 V)*(6.0 min) = 10.8 kJ 27-2 Chapter 27: Circuits Problem 4 A standard flashlight battery can deliver about 2.0 W-h of energy before it runs down. (a) If a battery costs 80, what is the cost of operating a 100-W lamp for 8.0 h using batteries? (b) What is the cost if power provided by an electric utility company, at 12 cents per kW-h, is used? Register to View Answer$320. (b) 4.8. Solution (a) The total amount of energy required to operate a 200-W lamp for 8.0 h is U = P*t = (100 W)*(8 h) = 800 W-h. To obtain 800 W-h from flashlight batteries would require 400 batteries (n = 800 W-h/2.0 W-h). If each battery cost 80 cents then the total cost would be cost = 400($0.80) = $320. This is not a cheap way to light your home! (b) The cost from the power company at the rate of 6 cents per kw-h is cost = ($ 0.06/kw-h)*(800 W-h) = 4.8. But it is difficult to keep the power company in the glove compartment of your car. Problem 12 (a) In Fig. 27-29 what value must R have if the current in the circuit is to be 1.0 mA? Take E1 = 2.0 V, E2 = 3.0V, and r1 = r2 = 3.0 . (b) What is the rate at which thermal energy appears in R? Register to View Answer994 . (b) 9.94 10-4 W. E1 r1 i R r2 E2 Figure 27-29 Solution (a) We assume that the current is going counter-clockwise in the circuit and apply the loop rule. Traversing the loop in a clockwise sense beginning at the upper left corner or - ir2 + E 2 - E 1 - ir1 - iR = 0 R= 2 - 1 3.0V - 2.0V - r2 - r1 = - 3.0 - 3.0 = 994 . i 1.0mA (b) The power dissipated in R is P = i2 R = (1.0 mA)2 (994 ) = 9.94 10-4 W Chapter 27: Circuits 27-3 Problem 22 A solar cell generates a potential difference of 0.10 V when a 500 resistor is connected across it and a potential difference of 0.15 V when a 1000- resistor is substituted. What are (a) the internal resistance and (b) the emf of the solar cell? (c) The area of the cell is 5.0 cm2 and the rate per unit area at which it receives energy from light is 2.0 mW/cm2 . What is the efficiency of the cell for converting light energy to thermal energy in the 1000- external resistor? Register to View Answer1000 . (b) 300 mV. (c) 0.225 %. Solution The circuit for this problem is shown in Diagram 27-22. The solar cell is shown with its internal resistance, r, and the loop current, i, is assumed to be clockwise. (a) Applying the loop rule (traversing the loop clockwise) -VR - E - VRr/R = 0. Diagram 27-22 We have now written the loop equation in terms of the two unknowns, E and r, and the two givens, VR and R. When R is 500 VR is 0.10 V and the loop equation becomes -0.10 V - E - (0.10 V)r/(500 ) = 0. When R is 1000 VR is 0.15 V and the loop equation becomes -0.15 V - E - (0.15 V)r/(1000 ) = 0. (a) first or (b) Solving these two equations simultaneously for r by subtracting the second from the 0.05 V + (0.15 V)r/(1000 ) - (0.10 V)r/(500 ) = 0. r = 1000 . Substituting the result of part (a) into one of the loop equations yields E = 0.300 V. (c) The efficiency of the cell is the ratio of the electrical power out to the radiant power in. The power out is Solar Cell r E i Load Resistor 27-4 Pout = VR2 /R = (0.15 V)2 /(1000 ) = 22.5 W. The power in is the intensity times the area Pin = IA = (2.0 mW/cm2 )*(5.0 cm2 ) = 10.0 mW. Therefore the efficiency is e = Pout/Pin = (22.5 W)/(10.0 mW) = 0.225 %. Chapter 27: Circuits Problem 25 Four 18.0- resistors are connected in parallel across a 25.0-V ideal battery. What is the current through the battery? Answer 5.56 A Solution Since resistors in parallel with the battery will all have the same potential difference across them and the magnitude of this potential difference is equal to the emf of the battery. The current through the battery is the sum of the currents through the four resistors. This is ibattery = 4iR = 4 25.0V = 4 = 5.56A . R 18.0 Problem 27 In Fig. 27-40, R1 = 100 , and R2 = 50 and the ideal batteries have emfs E 1 = 6.0 V, E 2 = 5.0 V, and E 3 = 4.0 V. Find (a) the current in resistor 1, (b) the current in each resistor 2, and the potential difference between points a and b. Register to View Answer50 mA; (b) 60 mA; (c) 9.0 V. E1 E2 a R1 E3 b R2 Figure 27-40 Chapter 27: Circuits Solution Referring to Diagram 27-27 we can write two loop equations for the circuit. c E1 E2 a R1 f i1 Diagram 27-27 (a,b) or For loop cdbeac: - E 1 + i2 R2 + E 3 + E 2 = 0. i2 = 27-5 d R2 i2 e E3 g b 1 - 2 - 3 6.0V - 5.0V - 4.0V = = -60mA . R2 50 The negative sign indicates that we assumed the wrong direction for i2 . Big deal! I did that on purpose. For loop aegfa: or - E 2 + i1 R1 = 0. i1 = 2 5.0V = = 50mA . R1 100 (c) The potential difference between points a and b is Vab = VB - VA = - E 2 - E 3 = - 5.0 V - 4.0 V = - 9.0 V. The negative sign indicates that point a is at a higher potential. 27-6 Problem 37 (a) Calculate the current through each ideal battery in Fig. 27-48. Assume that R1 = 1.0 , R2 = 2.0 , E 1 = 2.0 V, and E 2 = E 3 = 4.0 V. (b) Calculate Va Vb . Register to View AnswerLeft branch, 0.67 A, down; center branch, 0.33 A, up; right branch, 0.33 A, up. (b) 3.3 V. Chapter 27: Circuits R1 a R2 R1 E3 R1 E1 E2 b Figure 27-48 Solution Referring to Diagram 27-37a, we can write two loop equations and a junction equation. c R1 a R2 i2 E2 R1 R1 d E3 i1 E1 i3 f R1 b Diagram 27-37a e For loop cabfc: -i1 R1 - i2 R2 - E 2 - i1 R1 + E 1 = 0. For loop adeba: -i3 R1 - E 3 - i3 R1 - i3 R1 + E 2 + i2 R2 = 0. At junction a: i1 = i2 + i3 . (a) Solving these three equations simultaneously (Diagram 27-37b shows a Cramer's rule solution for i1 ) for the currents yields i1 = -2/3 A. i2 = -1/3 A. i3 = -1/3 A. Chapter 27: Circuits Here is how we would use Cramer's rule to solve for i1 : 27-7 1 -R2 0 2 - 3 R2 -2R1 0 -1 -1 R ( - 3 ) - ( 2R1 + R2 ) 1 i1 = = 2 2 -2R1 -R2 0 4R1 ( R2 + R1 ) 0 R2 -2R1 1 -1 -1 Diagram 27-37b 2 ( 4 V - 4 V ) - ( 2 + 2 ) 2V 2 =- 4 1 ( 2 + 1 ) 3 i1 = The negative signs indicate that I guessed wrong on the directions for all three currents! (b) We can calculate by any of the three paths between the two points. If we choose the path through E2 then R2 then we get from Diagram 27-37a VA - VB = E2 + i2 R2 = 4.0 V + (-1/3 A)*(2.0 ) = 3.3 V Note that if we still refer to the currents in the original diagram that we should use the results for the currents obtained from that diagram - i.e. we use -1/3 A and not +1/3 A. Problem 40 (a) In Fig. 27-51, R1 = 100 , R2 = R3 = 50 , R4 = 75 , and the ideal battery has emf E = 6.0 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resistance 2, (d) resistance 3, and (e) resistance 4? Register to View Answer119 . (b) i1 = 51 mA; i2 = i3 = 19 mA; i4 = 13 mA. R1 E R2 R4 R3 Figure 27-51 Solution (a) Resistances R2 , R3 and R4 are in parallel and may be replaced by an equivalent resistance, R2 3 4. 1 1 1 1 1 1 1 4 = + + = + + = R234 R2 R3 R4 50 50 75 75 R2 3 4 = 75/4 27-8 Chapter 27: Circuits This equivalent resistance is in series with R1 so that an equivalent resistance may replace all four Req = R1 + R2 3 4 Req = 100 + 18.75 = 118.75 . (b) The current through the equivalent resistance for the network is itot = E/Req = (6.0 V)/(118.75 ) = 50.5 mA. This will be the current through the seat of emf and therefore through R1 since it is in series with E. The potential difference across each of remaining three resistances is V = E - itotR1 = 6.0 V - (50.5 mA)*(100 ) = 0.947 V. Therefore the current through each of these resistors will be i2 = V/R2 = (0.947 V)/(50 ) = 18.9 mA, i3 = V/R3 = (0.947 V)/(50 ) = 18.9 mA, i4 = V/R4 = (0.947 V)/(75 ) = 12.6 mA. Problem 50 A simple ohmmeter is made by connecting a 1.50 V flashlight battery in series with a resistance R and an ammeter that reads from 0 to 1.00 mA, as shown in Fig. 27-56. Resistance R is adjusted so that when the clip leads are shorted together, the meter deflects to its full-scale value of 1.00 mA. What external resistance across the leads results in a deflection of (a) 10%, (b) 50%, and (c) 90% of full scale? (d) If the ammeter has a resistance of 20.0 and the internal resistance of the battery is negligible, what is the value of R? Register to View Answer13.5 k; (b) 1.5 k; (c) 167 ; (d) 1.5 k Solution (a) Let the sum of R and the internal resistance of the battery be Rm. Applying the loop theorem to the completed circuit when the clips are joined gives or E iRm = 0 Rm = E/i = (1.50 V) / (1.00 mA) = 1.5 k If we connect an external resistance, Rx , between the clips and apply the loop theorem when the current is i = 0.10 mA (10% of full scale) we get 0-1.00 mA E R Figure 27-56 Chapter 27: Circuits E iRm iRx = 0 Rx = (E/i) Rm = [(1.50 V) / (0.10 mA)] - 1.5 k = 1.35 k (b) As in part (a) Rx = (E/i) Rm = [(1.50 V) / (0.50 mA)] - 1.5 k = 1.5 k (c) As in part (a) Rx = (E/i) Rm = [(1.50 V) / (0.90 mA)] - 1.5 k = 167 (d) Answered in part (a): 1.5 k. 27-9 or Problem 54 In Fig. 27-60, R1 = 2.00R, the ammeter resistance is zero and the battery is ideal. What multiple of E/R gives the current in the ammeter? Answer E/7R. E R1 A R Figure 27-60 R R Solution Diagram 27-54 shows the circuit with branch currents labeled. We could solve for IA by setting up six equations in the six unknown currents and solving them simultaneously. We can make a little quicker work of this though if we realize that points c and d are at the same potential since there is no resistance between them along the path through the ammeter. Likewise, points a and b are at the same potential and points e and f are at the same potential. Therefore we can write g I E R h 2R c a I 1 IA R A I3 e Diagram 27-54 R b I2 d I4 f or Vac = Vb d 2RI1 = RI2 and or 2I1 = I2 Vce = Vdf RI3 = RI4 (1) 27-10 and I 3 = I4 Chapter 27: Circuits (2) We can also see that the potential difference from a to c to e is equal to E: RI3 + 2RI1 = E We can also write the junction theorem at points c and d respectively: I1 = I3 + IA I4 = I2 + IA These five equations are easily solved for IA = E/7R. Problem 55 In Fig. 27-61, Rs is to be adjusted in value by moving the sliding contact across it until points a and b are brought to the same potential. (One tests for this condition by momentarily connecting a sensitive ammeter between a and b; if these points are at the same potential, the ammeter will not deflect.) Show that when this adjustment is made, the following relation holds: R R x = Rs 2 . R1 An unknown resistor (Rx ) can be measured in terms of standard (Rs) using this device, which is called a Wheatstone bridge. (4) (5) (3) a R1 R2 Rs b R0 E Rx Figure 27-61 Solution Let the current going through Rs and Rx be is x and the current going through R1 and R2 be i1 2. If a and b are at the same potential then the potential differences across R1 and Rs are equal i1 2R1 = is xRs and the potential differences across R2 and Rx are equal i1 2R2 = is xRx . Dividing the first equation by the second yields R1 /R2 = Rs/Rx Chapter 27: Circuits or 27-11 R R x = Rs 2 . R1 Problem 56 In Figure a 27-62, voltmeter of resistance RV = 300 and an ammeter of resistance RA = 3.00 are being used to measure a resistance R in a circuit that also contains a resistance R0 = 100 and an ideal battery of emf E = 12.0 V. Resistance R is given by R = V/i, where V is the voltmeter reading and i is the current in resistance R. However, the ammeter reading is not i but rather i', which is i plus the current through the voltmeter. Thus, the ratio of the two meter readings is not R but only an apparent resistance R' = V/i'. If R = 85.0 W, what are (a) the ammeter reading, (b) the voltmeter reading, and (c) R'? (d) If RV is increased, does the difference between R' and R increase, decrease, or remain the same? Register to View Answer32.5 mA; (b) ; (c) decrease - as R v , R R R i iV V 300 E 3 A i' R0 Figure 27-62 Solution (a) Writing the loop theorems for the big outside loop and the small inner loop and the junction theorem gives us E iR i'RA i'R0 = 0 iR + iVRV = 0 i' = i + iV Solving for i' yields i' = 12.0 V = = 32.5 mA RV R 300 85 + 3.0 + 300 R + R + RA + R0 300 + 85 V (b) The voltmeter reading will be equal to the product iR. Solving the equations for i we get i = 25.3 mA and therefore V = iR = (25.3 mA)*(85.0 ) = 2.15 V (c) The apparent resistance is R' = V/I' = (2.15 V) / (32.5 mA) = 66.2 . 27-12 Chapter 27: Circuits (d) As RV is increased there will be less current in the voltmeter and therefore the difference between i and i' will decrease thus decreasing the difference between R and R'. If we divide the junction theorem equation by V we get i ' i iV = + V V V 1 1 1 = + R' R RV or Therefore as RV increases R' approaches R. Problem 58 A capacitor with initial charge q0 is discharged through a resistor. In terms of the time constant , how long is required for the capacitor to lose (a) the first one-third of its charge and (b) two-thirds of its charge? Register to View Answer; (b) Solution (a) The charge on a discharging capacitor is q = q0 e-t/. The charge remaining on the capacitor after losing a third of its charge will be 2q0 /3. 2 q0 = q0 e -t / 3 t 2 - = ln 3 3 t = ln = 0.405 2 (b) As in part (a) 3 t = ln = 1.10 1 Problem 60 In an RC series circuit E = 12 V, R = 1.4 M, and C = 1.8 F. (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 16 C? Register to View Answer2.52 s. (b) 21.6 C. (c) 3.40 s. Chapter 27: Circuits Solution (a) The capacitive time constant is = RC = (1.4 M)*(1.8 F) = 2.52 s. 27-13 (b) The maximum charge is reached when the current becomes zero and the potential difference across the capacitor is E. Therefore qmax = EC = (12 V)*(1.8 F) = 21.6 C. (c) The charge on the capacitor at time t is q(t) = qmax(1 - e-t/RC). Solving for t yields q(t) t = -RC ln 1 - . qmax Therefore the time at which q(t) = 16 C is 16 C t = -2.52s ln 1 - = 3.40s . 21.6 C Problem 62 A capacitor with an initial potential difference of 100 V is discharged through a resistor when a switch between them is closed at t = 0. At t = 10 s, the potential difference across the capacitor is 1.0 V. (a) What is the time constant of the circuit? (b) What is the potential difference across the capacitor at t = 17 s? Register to View Answer2.17 s. (b) 39.8 mV. Solution (a) The charge on the discharging capacitor at time t is q(t) = q0 e-t/RC. Therefore the potential difference across the capacitor at time t is V (t ) = q ( t ) qo -t RC = e = Voe-t RC . C C Solving for the capacitive time constant = RC = -t -10s = = 2.17s . ln (V ( t ) Vo ) ln (1.0V 100V ) 27-14 (b) The potential at time 17 s will be V(17 s) = V0 e-t/RC = (100 V)e-(17 s/2.17 s) = 39.8 mV. Chapter 27: Circuits Problem 65 In the circuit of Fig. 27-64, E = 1.2 kV, C = 6.5 F, R1 = R2 = R3 = 0.73 M. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are (a) current i1 in resistor 1, (b) current i2 in resistor 2, (c) current i3 in resistor 3? At t = (that is, after many time constants), what are (d) i1 , (e) i2 , and (f) i3 ? What is the potential difference V2 across resistor 2 at (g) t = 0 and (h) t = ? (i) Sketch V2 versus t between these two extreme times. R1 E S R2 C R3 Figure 27-64 Register to View Answer1.1 mA; (b) 0.55 mA; (c) 0.55 mA; (d) 0.82 mA; (e) 0.82 mA; (f) 0; (g) 400 V; (h) 600 V Solution (a) At time t = 0 the capacitor is uncharged therefore the potential difference across the capacitor is zero and, at that instant, the circuit behaves as if the capacitor were not present. The circuit is then just a resistor network connected to the seat of emf. Resistors R2 and R3 are in parallel with an equivalent resistance of 0.365 M. This 0.365-M equivalent resistance is in series with R1 so that the total equivalent resistance connected to the seat of emf is 1.095 M. The current through the battery, and therefore resistor R1 , is i= Requiv = 1.2kV = 1.1mA 1.095M (b) At time t = 0 the capacitor is uncharged and therefore has zero potential difference across it. Therefore the potential differences across R2 and R3 are equal which means that they must have equal currents since they have equal resistances. Since i2 and i3 are equal and their sum must be equal to i which means the two currents must each be 0.55 mA. (c) See part (b) (d) After a very long time the charge on the capacitor will no longer be changing and the current in resistor R3 will be zero. This means that the circuit essentially consists of the battery in series with resistors R1 and R2 . Thus there is 2*0.73 connected to the 1.2-kV battery with a resulting current of 0.82 mA through the battery. Chapter 27: Circuits (e) (f) (g) (h) Same as R1 , 0.82 mA. See part (d). Zero. See part (d). The potential difference is the product of the current, 400 V. The potential difference is the product of the current, 600 V. 27-15 Problem 90 In Fig. 28-8a calculate the potential difference between a and c by considering a path that contains R, r1 and E1 . Answer 2.5 V. Solution The potential difference between points a and c is VC - VA = -E1 + ir1 + iR VC - VA = -E1 + i(r1 + R). Battery 1 r1 E1 E2 r2 Figure 28-8a From Sample Problem 1 the current is i = 0.2396 A. Therefore VC - VA = -4.4 V + (0.2396 A)*(2.3 + 5.5 ) = -2.53 V. Problem xx Two light bulbs, one of resistance R1 and the other of resistance , where R1 > R2 , are connected to a battery (a) in parallel and (b) in series. Which bulb is brighter (dissipates more energy) in each case? Register to View AnswerR2 . (b) R1 . Solution The brighter bulb will be the one with the larger power. (a) If the bulbs are in parallel then they will have the same potential, V, difference across them and the power for each is and P1 = V2 /R1 P2 = V2 /R2 . Since R2 < R1 we have P2 > P1 . (b) If the bulbs are in series then they will have the same current, i, and the power in each will be 27-16 Chapter 27: Circuits and P1 = i2 R1 P2 = i2 R2 . In this case P1 > P2 since R1 > R2 . Problem xx (a) How much work does an ideal battery with a 12-V emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If 3.4 101 8 electrons pass through each second, what is the power of the battery? Register to View Answer1.92 10-18 J (= 12eV). (b) 6.53 W. Solution (a) The work done by the battery on the electron is W = qV = (-1.6 10-19 C)*(-12 V) = 1.92 10-18 J or W = qV = (-e)*(-12 V) = 12 eV. (b) The total amount of work done on the 3.4 101 8 electrons is W = (3.4 101 8)*(1.92 10-18 ) = 6.53 J. The power is therefore P = W/t = (6.53 J)/(1 s) = 6.53 W. Problem xx In Fig. 28-34, find the equivalent resistance between points (a) A and B, (b) A and C, and (c) B and C. Register to View Answer6.67 . (b) 6.67 . (c) zero. 20 20 Figure 28-34 20 Solution (a) Between points A and B the three resistors are in parallel therefore or 1/Req = 1/20 + 1/20 + 1/20 = 3/20 . Req = 20/3 . Chapter 27: Circuits (b) Like part (a) the three resistors are in parallel between points A and C. 27-17 (c) Points B and C are always at the same potential (they are connected by a zeroresistance conductor) therefore the resistance between B and C is zero. Problem xx In Fig. 28-12 assume that E = 5.0 V, r = 2.0 , R1 = 5.0 , and R2 = 4.0 . If the ammeter resistance RA = 0.10 , what percent error is made in reading the current? Assume that the voltmeter is not present. i E r a A b d c R2 R1 V Answer 0.9%. Solution The error is the difference between the current without the ammeter and the current with the ammeter. Without the ammeter the current would be i = E/(r + R1 + R2 ) i = (5.0 V)/(2.0 + 5.0 + 4.0 ) = 0.454 A. The current measured with the ammeter in place is imeasured = E/(r + R1 + R2 + RA) imeasured = (5.0 V)/(2.0 + 5.0 + 4.0 + 0.10 ) = 0.450 A. The percent error is (100%)*(i - imeasured)/i = (100%)*(0.454 A - 0.450 A)/(0.454 A) (100%)*(i - imeasured)/i = 0.9% Problem xx An initially uncharged capacitor C is fully charged by a device of constant emf E connected in series with a resistor R. (a) Show that the final energy stored in the capacitor is half the energy supplied by the emf. (b) By direct integration of i2 R over the charging time, show that the thermal energy dissipated by the resistor is also half the energy supplied by the emf. Figure 28-12 27-18 Chapter 27: Circuits Solution (a) The final energy stored in the capacitor, when the potential difference across the capacitor is E, is Uf = C E 2/2. To get the energy supplied by the battery we can integrate the power U R = 0 P(t) dt The power supplied by the battery is P(t) = E i(t) where i(t) = (E /R)e-t/RC. Therefore Ub = 0 E -t / RC E2 E e dt = R R 0 0 e -t / RC dt Ub = -CE 2 e-t / RC = CE 2 Thus the energy stored in the capacitor is half of the energy supplied by the battery. (b) The other half of the energy supplied is dissipated by the resistor. Integrating the power dissipated in the resistor U R = 0 i 2 R dt The current is i(t) = (E /R)e-t/RC. Therefore the energy dissipated is UR = 0 E2 E -t / RC e Rdt = R R 2 0 e-2t / RC dt CE 2 -t / RC Ub = - e 2 = 0 1 CE 2 2 Problem xx In Fig. 28-22 E1 = 12 V and E2 = 8 V. What is the direction of the current in the resistor? Which battery is doing positive work? Which point, A or B, is at the higher potential? Answer From B to A; #1; B. E1 E2 A R Figure 28-22 B Chapter 27: Circuits Solution (a) In this fairly simple case it may appear obvious that E1 will "overpower" E2 and the current will therefore go clockwise. If the circuit were much more complicated it might not be so obvious which way the current was going. In that case we could assume a current direction and apply the loop rule. For example we might assume that the current is going from A to B as in Diagram 28-4. 27-19 D E1 i A R Diagram 28-4 C E2 B If we now sum the potential differences starting from point A and proceeding through D, C, B and back to A we will get VA VA = E1 - E2 + iR or i = (E2 - E1 )/R = (8 V - 12 V)/R. The result for the magnitude of the current is going to turn out negative. This means that we assumed the wrong direction for the current and it must actually be going from B to A. Problem xx In Fig. 28-4a put E = 2.0 V and r = 100 . Plot (a) the current and (b) the potential difference across R, as functions of R over the range 0 to 500 . Make both plots on the same graph. (c) Make a third plot by multiplying together, for each value of R, the corresponding values on the two plotted curves. What is the physical significance of this third plot? E Solution (a) The current in resistor R is i = E/(r + R). The potential difference across resistor R is VR = iR = ER/(r + R). The graphs of these functions of R are shown in Diagram 28-8a. (b) The plot of the product of i and VR as a function of R is shown in Diagram 29-7b. Figure 28-4a 27-20 Chapter 27: Circuits Diagram 28-8 This graph is a graph of the power dissipated by R as a function of R. Note that the power dissipated has a maximum when R equals 100 that is also the value of the internal resistance of the seat of emf. ... View Full Document

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