ENGR HOMEWORK6SOLUTIONS


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1 ENGR 215 HOMEWORK6SOLUTIONS 1. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Prolem Prolem IDENTIFY: The equivlent resistnce will vry for the different connections ecuse the seriesprllel comintions vry, nd hence the current will vry. SET UP: First clculte the equivlent resistnce using the seriesprllel formuls, then use Ohm s lw ( V RI ) to find the current. EXECUTE: () 1/ R 1/(15. ) 1/(3. ) gives R 1.. I V / R (35. V)/(1. ) 3.5 A. () 1/ R 1/(1. ) 1/(35. ) gives R I (35. V)/(7.78 ) 4.5 A. (c) 1/ R 1/(2. ) 1/(25. ) gives R 11.11, so I (35. V)/(11.11 ) 3.15 A. (d) From prt (), the resistnce of the tringle lone is Adding the 3. internl resistnce of the ttery gives n equivlent resistnce for the circuit of Therefore the current is I (35.V)/(1.78 ) 3.25 A. EVALUATE: It mkes ig difference how the tringle is connected to the ttery IDENTIFY: The potentil drop is the sme cross the resistors in prllel, nd the current into the prllel comintion is the sme s the current through the 45. resistor. () SET UP: Apply Ohm s lw in the prllel rnch to find the current through the 45. resistor. Then pply Ohm s lw to the 45. resistor to find the potentil drop cross it. EXECUTE: The potentil drop cross the 25. resistor is V25 (25. )(1.25 A) V. The potentil drop cross ech of the prllel rnches is V. For the 15. resistor: I15 (31.25V)/(15. ) 2.83 A. The resistnce of the comintion is 25., so the current through it must e the sme s the current through the upper 25. resistor: I A. The sum of currents in the prllel rnch will e the current through the 45. resistor. ITotl 1.25 A 2.83 A 1.25 A 4.58 A Apply Ohm s lw to the 45. resistor: V45 (4.58 A)(45. ) 26 V () SET UP: First find the equivlent resistnce of the circuit nd then pply Ohm s lw to it. EXECUTE: The resistnce of the prllel rnch is 1/ R 1/(25. ) 1/(15. ) 1/(25. ), so R The equivlent resistnce of the circuit is Ohm s lw gives VBt (86.62 )(4.58 A) 398 V. EVALUATE: The emf of the ttery is the sum of the potentil drops cross ech of the three segments (prllel rnch nd two series resistors) () IDENTIFY: With the switch open, the circuit cn e solved using seriesprllel reduction. SET UP: Find the current through the unknown ttery using Ohm s lw. Then use the equivlent resistnce of the circuit to find the emf of the ttery.
2 EXECUTE: The 3. nd 5. resistors re in series, nd hence hve the sme current. Using Ohm s lw I (15. V)/(5. ).3 A I. The potentil drop cross the 75. resistor is the sme s the 5 3 potentil drop cross the 8. series comintion. We cn use this fct to find the current through the 75. resistor using Ohm s lw: V75 V8 (.3 A)(8. ) 24. V nd I (24. V)/(75. ).32 A. 75 The current through the unknown ttery is the sum of the two currents we just found: ITotl.3 A.32 A.62 A The equivlent resistnce of the resistors in prllel is 1/ Rp =1/(75. ) 1/(8. ). This gives Rp The equivlent resistnce seen y the ttery is Requiv Applying Ohm s lw to the ttery gives Requiv ITotl (58.7 )(.62 A) 36.4 V. () IDENTIFY: With the switch closed, the 25.V ttery is connected cross the 5. resistor. SET UP: Tke loop round the right prt of the circuit. EXECUTE: Ohm s lw gives I (25. V)/(5. ).5 A. EVALUATE: The current through the 5. resistor, nd the rest of the circuit, depends on whether or not the switch is open IDENTIFY: We need to use Kirchhoff s rules. SET UP: Tke loop round the outside of the circuit, use the current t the upper junction, nd then tke loop round the right side of the circuit. EXECUTE: The outside loop gives 75. V (12. )(1.5 A) (48. ) I48, so I A. At junction we hve 1.5A I A, nd I.313 A. A loop round the right prt of the circuit gives (48 )(1.188 A) (15. )(.313 A) V, with the polrity shown in the figure in the prolem. EVALUATE: The unknown ttery hs smller emf thn the known one, so the current through it goes ginst its polrity IDENTIFY: The glvnometer is represented in the circuit s resistnce R c. Use the junction rule to relte the current through the glvnometer nd the current through the shunt resistor. The voltge drop cross ech prllel pth is the sme; use this to write n eqution for the resistnce R. SET UP: The circuit is sketched in Figure Figure We wnt tht I 2 A in the externl circuit to produce Ifs 224 A through the glvnometer coil. EXECUTE: Applying the junction rule to point gives I Ifs Ish. Ish I Ifs 2 A 224 A A The potentil difference V etween points nd must e the sme for oth pths etween these two points: I ( R R ) I R fs c sh sh I R (19 98 A)( 25 ) A sh sh R Ifs Rc EVALUATE: R sh R R c ; current tht goes through R c. most of the current goes through the shunt. Adding R decreses the frction of the
3 26.4.IDENTIFY: An unchrged cpcitor is plced into circuit. Apply the loop rule t ech time. SET UP: The voltge cross cpcitor is V q/ C. C EXECUTE: () At the instnt the circuit is completed, there is no voltge cross the cpcitor, since it hs no chrge stored. () Since VC, the full ttery voltge ppers cross the resistor VR 245 V. (c) There is no chrge on the cpcitor. 245 V (d) The current through the resistor is i 327 A. R 75 totl (e) After long time hs pssed the full ttery voltge is cross the cpcitor nd i. The voltge cross the cpcitor lnces the emf: VC 245 V. The voltge cross the resistor is zero. The cpcitor s chrge is 6 3 q CV C (4 6 1 F) (245 V) C. The current in the circuit is zero. EVALUATE: The current in the circuit strts t.327 A nd decys to zero. The chrge on the cpcitor strts 3 t zero nd rises to q C IDENTIFY: The cpcitors, which re in prllel, will dischrge exponentilly through the resistors. SET UP: Since V is proportionl to Q, V must oey the sme exponentil eqution s Q,. V V e The current is I ( V / R) e. EXECUTE: () Solve for time when the potentil cross ech cpcitor is 1. V: () t RC ln( V / V ) (8. )(35. µ F) ln(1/45) 421 µ s 4.21 ms I ( V / R) e. Using the ove vlues, with V 45. V, gives I.125 A. EVALUATE: Since the current nd the potentil oth oey the sme exponentil eqution, they re oth reduced y the sme fctor (.222) in 4.21 ms IDENTIFY nd SET UP: Ohm s lw nd Eq. (25.18) cn e used to clculte I nd P given V nd R. Use Eq. (25.12) to clculte the resistnce t the higher temperture. () EXECUTE: When the heter element is first turned on it is t room temperture nd hs resistnce R 2. V 12 V I 6 A R V (12 V) P 72 W R 2 () Find the resistnce R(T) of the element t the operting temperture of 28 C. Tke T 23 C nd R 2. Eq. (25.12) gives 3 1 R( T) R (1 ( T T )) 2 (1 (2 8 1 (C ) )(28 C 23 C)) V 12 V I 3 5 A R V (12 V) P 42 W R IDENTIFY: Apply the junction rule nd the loop rule to the circuit. SET UP: Becuse of the polrity of ech emf, the current in the 7  resistor must e in the direction shown in Figure Let I e the current in the 24.V ttery. EXECUTE: The loop rule pplied to loop (1) gives: 24 V (1 8 A)(7 ) I(3 ). I 3 8 A. The junction rule then sys tht the current in the middle rnch is 2. A, s shown in Figure The loop rule pplied to loop (2) gives: (1 8 A)(7 ) (2 A)(2 ) nd 8 6 V. EVALUATE: We cn check our results y pplying the loop rule to loop (3) in Figure 26.64: 24 V (2 A)(2 ) (3 8 A)(3 ) nd 24 V 4 V 11 4 V 8 6 V, which grees with our result from loop (2).
4 Figure () IDENTIFY: Brek the circuit etween points nd mens no current in the middle rnch tht contins the 3. resistor nd the 1.V ttery. The circuit therefore hs single current pth. Find the current, so tht potentil drops cross the resistors cn e clculted. Clculte V y trveling from to, keeping trck of the potentil chnges long the pth tken. SET UP: The circuit is sketched in Figure Figure EXECUTE: Apply the loop rule to loop (1). 12 V I( ) 8 V I(2 1 ) 12 V 8 V I 4444 A. 9 To find V strt t point nd trvel to, dding up the potentil rises nd drops. Trvel on pth (2) shown on the digrm. The 1  nd 3  resistors in the middle rnch hve no current through them nd hence no voltge cross them. Therefore, V 1 V 12 V I(1 1 2 ) V ; thus V V 2 V ( 4444 A)(4 ) 22 V (point is t higher potentil) EVALUATE: As check on this clcultion we lso compute V y trveling from to on pth (3). V 1 V 8 V I(2 1 2 ) V V 2 V ( 4444 A)(5 ) 22 V, which checks. () IDENTIFY nd SET UP: With points nd connected y wire there re three current rnches, s shown in Figure
5 Figure The junction rule hs een used to write the third current (in the 8.V ttery) in terms of the other currents. Apply the loop rule to loops (1) nd (2) to otin two equtions for the two unknowns I1 nd I 2. EXECUTE: Apply the loop rule to loop (1). 12 V I (1 ) I (2 ) I (1 ) 1 V I (3 ) I (1 ) V I (4 ) I (4 ) (2 ) I (2 ) I 1 V eq. (1) Apply the loop rule to loop (2). ( I I )(2 ) ( I I )(1 ) 8 V ( I I )(2 ) I (3 ) 1 V I (1 ) V (5 ) I (9 ) I eq. (2) Solve eq. (1) for I 2 nd use this to replce I 2 in eq. (2). I 5 A I V (5 ) I (9 )( 5 A I ) (14 ) I 6 5 V so I (6 5 V)/(14 ) 464 A. I2 5 A 464 A 36 A. The current in the 12.V ttery is I1 464 A. EVALUATE: We cn pply the loop rule to loop (3) s check. 12 V I (1 2 1 ) ( I I )(2 1 2 ) 8 V 4 V 1 86 V V, s it should.
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