Chapter 28. Direct-Current Circuits

Transcription

1 Chapter 28. Direct-Current Circuits esistors in Series and Parallel (gnore internal resistances for batteries in this section.) A 5- resistor is connected in series with a 3- resistor and a 16-V battery. What is the effective resistance and what is the current in the circuit? e = = +5 ; e = V 16 V 8 = 2.00 A 16 V A 15- resistor is connected in parallel with a 30- resistor and a 30-V source of emf. What is the effective resistance and what total current is delivered? e 1 2 (15 )(30 ) ; e = V V 30 V 10 ; = 3.00 A n Problem 28-2, what is the current in 15 and 30- resistors? 30 V For Parallel: V 15 = V 30 = 30 V; 15 ; = 2.00 A V ; = 1.00 A Note: = T = 3.00 A What is the equivalent resistance of 2, 4, and 6- resistors connected in parallel? e = + ; e =

2 28-5. An 18- resistor and a 9- resistor are first connected in parallel and then in series with a 24-V battery. What is the effective resistance for each connection? Neglecting internal resistance, what is the total current delivered by the battery in each case? e 1 2 (18 )(9 ) ; e = V 18 9 V 6.00 ; = 4.00 A e = = ; e = V 27 = A A 12- resistor and an 8- resistor are first connected in parallel and then in series with a 28-V source of emf. What is the effective resistance and total current in each case? e 1 2 (1)(8 ) ; e = V 28 V 4.80 ; = 5.83 A 28 V V 8 e = = 1 +8 ; e = 20.0 V 28 V 20 = 1.40 A An 8- resistor and a 3- resistor are first connected in parallel and then in series with a 12-V source. Find the effective resistance and total current for each connection? e ()(8 ) + 8 ; e = 2.18 V 12 V 2.18 ; = 5.50 A e = = +8 ; e = 11.0 V 12 V 11 = 1.09 A 386

3 28-8. Given three resistors of 80, 60, and 40, find their effective resistance when connected in series and when connected in parallel. Series: e = ; e = 180 Parallel: ; e = e i Three resistances of 4, 9, and 11 are connected first in series and then in parallel. Find the effective resistance for each connection. Series: e = ; e = 24.0 Parallel: ; e = e i * A 9- resistor is connected in series with two parallel resistors of 6 and 1. What is the terminal potential difference if the total current from the battery is 4 A? ()(1) e ; e + 9 = V T = = (4 A)(1); V T = 52.0 V 1 4 A V T 9 * For the circuit described in Problem 28-10, what is the voltage across the 9- resistor and what is the current through the 6- resistor? V 9 = (4 A)(9 ) = 36 V; V 9 = 36.0 V The rest of the 52 V drops across each of the parallel resistors: V 6 = V 7 = 52 V 36 V; V 6 = 16 V 6 V V ; 6 = 2.67 A 387

4 * Find the equivalent resistance of the circuit drawn in Fig Start at far right and reduce circuit in steps: = = ; ()() '' ; e = ; e = * Find the equivalent resistance of the circuit shown in Fig Start at far right and reduce circuit in steps: = = ; ()() ' ; = + 3 = e (5 )() 2.2 ; e = e * f a potential difference of is applied to the circuit drawn in Fig , what is the current and voltage across the 1- resistor? e = A; 8 The voltage across the 3 and 6- parallel 1 connection is found from t and the 2- combination resistance: 6 V V 3 = V 6 = ()(3.00 A); V 6 = 6.00 V; 6 1 A Thus, 1 = 6 = 1.00 A, and V 1 = (1 A)(1 ) = 1 V; V 1 = 1 V; 1 = 1 A 388

5 * f a potential difference of 12-V is applied to the free ends in Fig , what is the current and voltage across the 2- resistor? 1 12 V e = A; V 12 V Note that V 5 = 12 V; A V V 3,6 = (2.4 A)() = 4.80 V; A = 1 = 1.60 A; V 2 = (1.6 A)() = 3.20 V = 1.60 A; V 2 = 3.20 V 12 V 5 12 V 12 V 2 EMF and Terminal Potential Difference A load resistance of 8 is connected in series with a 18-V battery whose internal resistance is 1.0. What current is delivered and what is the terminal voltage? E 18V ; = 2.00 A r L A resistance of is placed across a 12-V battery whose internal resistance is 0.. What is the current delivered to the circuit? What is the terminal potential difference? E 12V ; r 0. L = 1.90 A V T = E r = 12 V (1.90 A)(0.); V T = 11.4 V 389

6 Two resistors of 7 and 1 are connected in parallel with a 16-V battery whose internal resistance is What is the terminal potential difference and the current in delivered to the circuit? (7 )(1) ' 4.67 ; e = V E 16 V ; = 3.25 A V T = E r = 16 V (3.25 A)(0.25 ); r' V T = 15.2 V; = 3.25 A The open-circuit potential difference of a battery is 6 V. The current delivered to a 4- resistor is 1.40 A. What is the internal resistance? E = L + r; r = E - L E r L 6 V - (1.40 A)() ; r = A A dc motor draws 20 A from a 120-V dc line. f the internal resistance is 0., what is the terminal voltage of the motor? V T = E r = 120 V (20A)(0.); V T = 116 V For the motor in Problem 28-21, what is the electric power drawn from the line? What portion of this power is dissipated because of heat losses? What power is delivered by the motor? P i = E = (120 V)(20 A); P L = r = (20 A) 2 (0.2); P i = 2400 W P L = 80 W P o = V T = (116 V)(20 A); P o = 2320 W; Note: P i = P L + P o ; 2400 W = 80 W W 390

7 A 2- and a 6- resistor are connected in series with a 24-V battery of internal resistance 0.5. What is the terminal voltage and the power lost to internal resistance? e = = 8.50 ; E 2.82 A 8.5 e V T = E r = (2.82 A)(0.5 ); V T = 22.6 V P L = r = (2.82 A) 2 (0.5 ); P L = 3.99 W * Determine the total current and the current through each resistor for Fig when E =, 1 = 2 = 3 = 1 4 = and r = 0.. e 1,2 ()() ; 1,2,3 = + 1 = 3 ()() 1.20 ; 3 e = = = = T ; T = 15.0 A V V 4 = V 1.2 = (1.)(15 A) = 18 V = 9.0 A; 3 = 15 A 9 A = 6 A; V 3 = (6 A)(1 ) = 6 V; V 1 = V 2 = 18 V 6 V; 12 V 12 V V 1 = V 2 = 12 V; 4 A; 1 2 A ; T = 15 A, 1 = 2 A, = 3 = 4 = 9 A The solution is easier using Kirchhoff s laws, developed later in this chapter. 391

8 * Find the total current and the current through each resistor for Fig when E = 50 V, 1 = 1 2 = 3 = 4 = 8 and r = 0.. 1,2 (1)() ; 1,2,3 = + = (10 )(8 ) e = = V T ; 4.8 T = 10.3 A 3 = 45.9 V V 4 = V p = (4.4)(10.3 A) = 45.9 V = V V V V 4 = 5.73 A; 3 = 10.3 A 5.73 A = 4.59 A; V 3 = (4.59 A)() = 27.5 V; 18.4 V 18.4 V V 1 = V 2 = 45.9 V 27.5 V = 18.4 V; 3.06 A; A; 1 T = 10.3 A, 1 = 1.53 A, = = = 5.73 A Kirchhoff s Laws Apply Kirchhoff s second rule to the current loop in Fig What is the net voltage around the loop? What is the net drop? What is the current in the loop? ndicate output directions of emf s, assume direction of 20 V + current, and trace in a clockwise direction for loop rule: E = ; 20 V 4 V = () + (); (8 ) = 16 V; 16 V 8 ; = 2.00 A 4 V Net voltage drop = E = 16 V; = (8 )(2 A) = 16 V 392

9 Answer the same questions for Problem where the polarity of the 20-V battery is changed, that is, its output direction is now to the left? ( efer to Fig. in Prob ) E = -20 V 4 V = -; = () + ( = (8) E = ; - = (8 ); = A; = (8 )(-4 A) = - The minus sign means the current is counterclockwise (against the assume direction) * Use Kirchhoff s laws to solve for the currents through the circuit shown as Fig First law at point P: 1 + = 3 Current rule 1 Loop A (2 nd law): E = Loop rule 5 V 4 V = () 1 + () 1 () P A 5 V Simplifying we obtain: (1) = 1 A 4 V B 3 1 Loop B: 4 V 3 V = () + () 3 + (1 ) 3 Simplifying: (2) = 1 A, but 3 = V Substituting we have: 6 + 4( 1 + ) = 1 A or (3) = 1 A From which, 1 = 0.25 A 2.5 ; Substituting into (1): 6(0.25 A 2.5 ) 6 = 1 A 1.5 A 15 6 = 1 A; -21 = -0.5 A; = A; = 23.8 ma Putting this into (1), we have: 6 1 6( A) = 1 A, and 1 = 190 ma Now, 1 + = 3 so that 3 = 23.8 ma ma or 3 = 214 ma * Use Kirchhoff s laws to solve for the currents in Fig Current rule: = or (1) 3 = 1 Loop A: 20 V = () 1 + () ; (2) = 20 A Loop B: 8 V = () 3 + () ; (3) = 4 A Outside Loop: 20 V 8 V = () () or = 4 A 20 V P 1 4 V A B 8 V 5 V 3 393

10 * (Cont.) (3) = 4 A and 3 = 1 3( 1 ) + 2 = 4 A; 3 1 = 5 4 A (2) = 20 A; (5 4 A) + 4 = 20; = 2.67 A; 3 1 = 5(2.67 A) 4 A; 1 = 3.11 A 3 = 1 = 2.67 A 3.11 A = A 20 V P 1 4 V A B 5 V 3 Note: 3 goes in opposite direction to that assumed. 1 = 3.11 A, = 2.67 A, 3 = A 8 V * Apply Kirchhoff s laws to the circuit of Fig Find the currents in each branch. Current rule: (1) = V Applying loop rule gives six possible equations: 1 (2) = 3 A; (3) = (4) = 6A; (5) = -3A (6) = 6 A (7) = 3A 6 V 4 Put 4 = ; into (4): ( ) = 6 A = 6 A Now, solving (2) for 1 gives: 1 = 2 A 2, which can be used in the above equation. -6(2 A 2 ) = 6 A, which gives: = 18 A But, from (3), we put = into above equation to find that: 3 = A From (2): (0.439 A) = 3 A; and 1 = A From (3): 3 5(0.439 A) = 0; and = A From (4): 5(0.439 A) = 6 A; and 4 = A Currents in each branch are: 1 = 536 ma, = 732 ma, 3 = 439 ma, 4 = 634 ma Note: Not all of the equations are independent. Elimination of two may yield another. t is best to start with the current rule, and use it to eliminate one of the currents quickly. 394

11 The Wheatstone Bridge A Wheatstone bridge is used to measure the resistance x of a coil of wire. The resistance box is adjusted for, and the contact key is positioned at the 45 cm mark when measured from point A of Fig Find x. ( Note: l 1 + l 2 = 100 cm ) x l l 32 ; x = ()(55 cm) (45 cm) Commercially available Wheatstone bridges are portable and have a self-contained galvanometer. The ratio 2 / 1 can be set at any integral power of ten between and 1000 by a single dual switch. When this ratio is set to 100 and the known resistance is adjusted to 46.7, the galvanometer current is zero. What is the unknown resistance? x 2 3 (46.7 )(100) ; x = n a commercial Wheatstone bridge, 1 and 2 have the resistances of 20 and 40, respectively. f the resistance x is 1, what must be the known resistance 3 for zero galvanometer deflection? 2 20 x 3 (1) 1 40 ; x = 7.00 Challenge Problems: esistances of 3, 6, and 9 are first connected in series and then in parallel with an 36-V source of potential difference. Neglecting internal resistance, what is the current leaving the positive terminal of the battery? e = i = = 18; 36 V 18 ; = 2.00 A 395

12 (Cont.) ; e = 1.6; 9 e i 36 V 1.6 ; = 22.0 A Three 3- resistors are connected in parallel. This combination is then placed in series with another 3- resistor. What is the equivalent resistance? ; ' e = + V e = 1 + ; e = * Three resistors of 4, 8, and 1 are connected in series with a battery. A switch allows the battery to be connected or disconnected from the circuit? When the switch is open, a voltmeter across the terminals of the battery reads 50 V. When the switch is closed, the voltmeter reads 48 V. What is the internal resistance in the battery? L = = 2; E = 50 V; V T = 48 V = L 48 V 2.00 A ; E V T = r; 50 V 48 V = r 50 V - 48 V r ; r = A * The generator in Fig develops an emf of E1 = and has an internal resistance of 0.. The generator is used to charge a battery E2 = 12 V whose internal resistance is 0.. Assume that 1 = and 2 =. What is the terminal voltage across the generator? What is the terminal voltage across the battery? - 12 V 1.14 A V 1 = E1 r = (1.14 A)(0.) = 23.8 V; V 2 = 12 V + (1.14 A)(0.) = 12.3 V r r 12 V 396

13 * What is the power consumed in charging the battery for Problem Show that the power delivered by the generator is equal to the power losses due to resistance and the power consumed in charging the battery. P = E = ()(1.143 A) P e = W P = e = (1.143 A) 2 (10.5 ); P = W r r 12 V P V = (12 V)(1.143 A) = W; P e = P + P V ; 27.4 W = 13.7 W W * Assume the following values for the parameters of the circuit illustrated in Fig. 28-8: E1 = 100 V, E2 = 20 V, r 1 = 0.. r 2 = 0., and = What are the terminal voltages V 1 and V 2? What is the power lost through the 4- resistor? 100 V - 20 V 17.0 A 100 V V 1 = E1 r = 100 V (17.0 A)(0.) = 94.9 V; r r 20 V V 2 = 20 V + (17.0 A)(0.) = 26.8 V; P = = (17 A) 2 () = 1160 W * Solve for the currents in each branch for Fig V Current rule: 1 = + 3 ; Loops: E = s (1) = 12 A; (2) = 6 A (2) = 3 A; (3) = 18 A From (1): 5( + 3 ) + 10 = 12 A = 12 A P V A 10 B 3 Multiplying this equation by 2: = -24 A Now add this to (2): = -21 A and = A 6 V Now, from (1) and from (2): 1 = 1.20 A and 3 = A 397

14 * f the current in the 6- resistor of Fig is 2 A, what is the emf of the battery? Neglect internal resistance. What is the power loss through the 1- resistor? V 6 = (2 A)() = 12 V; V 3 = 12 V = 3 () 2A 1 12 V 3 4 A ; T = 2 A + 4 A = 6 A; 3 T E V 1 = (1 )(6 A) = 6 V; E = 6 V + 12 V = 18 V; P = T 2 = (6 A) 2 (1 ) = 36 W Critical Thinking Problems * A three-way light bulb uses two resistors, a 50-W filament and a 100-W filament. A three-way switch allows each to be connected in series and provides a third possibility by connecting the two filaments in parallel? Draw a possible arrangement of switches than will accomplish these tasks. Assume that the household voltage is 120 V. What are the resistances of each filament? What is the power of the parallel combination? Switch can be set at A, B, or C to give three possibilities: V (120 V) P ; 1 = 288 ; 50 W 2 (120 V) 2 = W 1 2 B T 120 V For Parallel: e (288 )(14) A C V P (120 V) 9 ; P = 150 W

15 * The circuit illustrated in Fig consists of a 12-V battery, a 4- resistor, and a switch. When new, the internal resistance of the battery is 0., and a voltmeter is placed across the terminals of the battery. What will be the reading of the voltmeter when the switch is open and when it is closed? After a long period of time, the experiment is repeated and it is noted that open circuit reading is unchanged, but the terminal voltage has reduced by 10 percent. How do you explain the lower terminal voltage? What is the internal resistance of the old battery? When new: E 12 V ; r + 0. = 2.73 A V T = E r = 12 V (2.73 A)(0.); V T = 10.9 V V T reduced by 10% due to increase of r int. V = 10.9 V 0.1(10.9 V); V = 9.81 V V = L = 9.81 V; 9.81 V E V ' 2.45 A; V ' E r; r E V ' 12 V V r ;.45 A r = 0.89 * Given three resistors of 3, 9, and 18 list all the possible equivalent resistances that can be obtained through various connections? All in parallel: ; e = e All in series: e = = ; e = 30 Parallel (3,9) in series with (18): Parallel (3,18) in series with (9): Parallel (9,18) in series with (3): e e e ()(9 ) 18 ; + 9 ()(18 ) 9 ; + 18 (9 )(18 ) ; e = 20. e = 11. e =

16 *28-43 (Cont.) Series (3 + 9) in parallel with (18): Series (3 + 18) in parallel with (9): Series (9 + 18) in parallel with (3): e e e (1)(18 ) ; (9 )(21 ) ; ()(27 ) ; + 27 e = 20. e = 6.30 e = 2.70 * efer to Fig , assume that E =, 1 = 8 2 = 3 = 4 = and r = 0.5. What current is delivered to the circuit by the 24-V battery? What are the voltage and current for the 8- resistor? 1,2 e ()(8 ) 2.18 ; 1,2,3 = = (4.18 )() 2.0; 4.18 e = = = = T ; 2.5 T = 9.43 A V 4 = V 1,2,3 = (9.43 A) (2.0) = 19.3 V V A ; 3 = 9.43 A 4.82 A = 4.61 A; V 3 = (4.61 A)() = 9.23 V 10.1 V V 1 = V 2 = 19.3 V 9.23 V; V 1 = V 2 = 10.1 V; A ; 8 Finally, for the 8- resistor: V 1 = 10.1 V and 1 = 1.26 A 400

17 * What is the effective resistance of the external circuit for Fig if internal resistance is neglected. What is the current through the 1- resistor? ,5 ()(8 ) 2.18 ; 3,4,5 = = (3.18 )() 1,2,3,4 = 3.18 ; 1,2,3,4 = e = ; e = in the 1- resistor is same as in 3,4,5 T 3.95 A 6.08 V 2,3,4,5 = (3.95 A)(2.08 ) = 8.21 V; Also V 3,4,5 = 8.21 V 8.21 V 3,4, A; 3.18 Therefore 3 = 2.58 A in 1- resistor e = 6.08 ; 3 = 2.58 A 401