( ) Homework Solutions Physics 8B Spring 09 Chpt. 32 5,18,25,27,36,42,51,57,61,76

Transcription

1 Homework Solutons Physcs 8B Sprng 09 Chpt. 32 5,8,25,27,3,42,5,57,, Model: Assume deal connectng wres and an deal battery for whch V bat = E. Please refer to Fgure EX32.5. We wll choose a clockwse drecton for. Note that the choce of the current s drecton s arbtrary because, wth two batteres, we may not be sure of the actual current drecton. The 3 V battery wll be labeled and the V battery wll be labeled 2. Solve: (a) Krchhoff s loop law, gong clockwse from the negatve termnal of the 3-V battery s ( ) V = V = V + V + V = closed loop bat R bat V +3 V (8 Ω) + V = 0 = = 0.5 A 8 Ω Thus, the current through the 8 Ω resstor s 0.5 A. Because s postve, the current s left to rght (.e., clockwse). (b) Assess: The graph shows a 3 V gan n battery, a 9 V loss n the resstor, and a gan of V n battery 2. The fnal potental s the same as the ntal potental, as rured Model: Assume deal connectng wres but not an deal battery. The crcut for an deal battery s the same as the crcut n Fgure EX32.8, except that the Ω resstor s not present. Solve: n the case of an deal battery, we have a battery wth E = 5 V connected to two seres resstors of 0 Ω and 20 Ω resstance. Because the uvalent resstance s R = 0 Ω + 20 Ω = 30 Ω and the potental dfference across R s 5 V, the current n the crcut s The potental dfference across the 20 Ω resstor s V E 5 V = = = = 0.50 A R R 30 Ω V 20 = R = (0.50 A)(20 Ω) = 0.0 V n the case of a real battery, we have a battery wth E = 5 V connected to three seres resstors: 0 Ω, 20 Ω, and an nternal resstance of.0 Ω. Now the uvalent resstance s R = 0 Ω + 20 Ω +.0 Ω = 3 Ω The potental dfference across R s the same as before ( E = 5 V). Thus, V E 5 V = = = = A R R 3 Ω Therefore, the potental dfference across the 20 Ω resstor s

2 ( )( ) V 20 = R = A 20 Ω = 9.8 V That s, the potental dfference across the 20 Ω resstor s reduced from 0.0 V to 9.8 V due to the nternal resstance of Ω of the battery. The percentage change n the potental dfference s 0.0 V 9.8 V 00 = 3.2% 0.0 V Model: The connectng wres are deal wth zero resstance. Solve: n the frst step, the resstors 00 Ω, 00 Ω, and 00 Ω n the top branch are n seres. Ther combned resstance s 300 Ω. n the mddle branch, the two resstors, each 00 Ω, are n seres. So, ther uvalent resstance s 200 Ω. n the second step, the three resstors are n parallel. Ther uvalent resstance s The uvalent resstance of the crcut s 54.5 Ω. R = 300 Ω Ω + 00 Ω R = 54.5 Ω Model: Groundng does not affect a crcut s behavor. Please refer to Fgure EX Solve: Because the earth has V earth = 0 V, pont d has a potental of zero. n gong from pont d to pont a, the potental ncreases by 9 V. Thus, pont a s at a potental of 9 V. Let us calculate the current n the crcut before calculatng the potentals at ponts b and c. Applyng Krchhoff s loop rule, startng clockwse from pont d, ( V) V9 V bat V2 Ω V V bat V = = 0 Ω 3 V +9 V (2 Ω) V ( Ω) = 0 = = A 3 Ω There s a drop n potental from pont a to pont b by an amount R = ( A)(2 Ω) = 2 V. Thus, the potental at pont b s 9 V 2 V = 7 V. The potental decreases from 7 V at pont b to 7 V V = V at pont c. There s a further decrease n potental across the Ω resstor of R = ( A)( Ω) = V. That s, the potental of V at c becomes 0 V at pont d, as t must. n summary, the potentals at a, b, c, and d are 9 V, 7 V, V, and 0 V Please refer to Fgure P32.3. Solve: Bulbs D and E are n seres, so the same current wll go through both and make them ually brght (D = E). Bulbs B and C are n parallel, so they have the same potental dfference across them. Because they are dentcal bulbs wth ual resstances, they wll have ual currents and be ually brght (B = C). Now the uvalent resstance of B + C n parallel s less than the resstance of E, so the total resstance along the path through A s less than the total resstance along path through D. The two paths have the same total potental dfference the emf of the battery so more current wll flow through the A path than through the D path. Consuently, A wll have more current than D and E and wll be brghter than D and E (A > D = E). Bulbs B and C each have half the current of A, because the current splts at the juncton, so A s also brghter than B and C (A > B = C). The remanng ssue s how B and C compare to D and E. Suppose B and C were replaced by wres wth zero resstance, leavng just bulb A n the mddle path. Then the resstance of the path through A would be half of the 2

3 resstance of the path through D. Ths would mean that the current through A would be twce the current through D, so A = 2 D. When B and C are present, ther resstance adds to the resstance of A to lower the current through the mddle path. So n realty, A < 2 D. We already know that B = C =, so we can conclude that 2 A B = C < D. Snce the current through B and C s less than the current through D and E, D and E are brghter than B and C. The fnal result of our analyss s A > D = E > B = C Model: Use the laws of seres and parallel resstances. Solve: Despte the dagonal orentaton of the 2 Ω resstor, the Ω, 2 Ω, and 4 Ω resstors are n parallel because they have a common connecton at both the top end and at the bottom end. Ther uvalent resstance s R = + + = 2 Ω Ω 2 Ω 4 Ω The trckest ssue s the 0 Ω resstor. t s n parallel wth a wre, whch s the same thng as a resstor wth R = 0 Ω. The uvalent resstance of 0 Ω n parallel wth 0 Ω s R = + = ( ) = = 0 Ω 0 Ω 0 Ω n other words, the wre s a short crcut around the 0 Ω, so all the current goes through the wre rather than the resstor. The 0 Ω resstor contrbutes nothng to the crcut. So the total crcut s uvalent to a 2 Ω resstor n seres wth the 2 Ω uvalent resstance n seres wth the fnal 3 Ω resstor. The uvalent resstance of these three seres resstors s R ab = 2 Ω + 2 Ω + 3 Ω = 7 Ω Model: Assume an deal battery and deal connectng wres. 3

4 Solve: (a) Groundng one pont doesn t affect the basc analyss of the crcut. n Fgure P32.5, there s a sngle loop wth a sngle current flowng n the clockwse drecton. Applyng Krchhoff s loop law clockwse from the lower rght corner gves V = V 4 2 = 0 V = 2 V = 0.50 A 24 Ω Knowng the current, we can use V = R to fnd the potental dfference across each resstor: V 8 = 4 V V 4 = 2 V V 2 = V The purpose of groundng one pont n the crcut s to establsh that pont as the specfc potental V = 0 V. Groundng pont d makes that potental there V d = 0 V. Then we can use the known potental dfferences to fnd the potental at other ponts n the crcut. Pont a s 4 V less than pont d (because potental decreases n the drecton of current flow), so V a = V d 4 V = 4 V. Pont b s 2 V more than pont a because of the battery. So V b = V a + 2 V = 8 V. Pont c s 2 V less than pont b, so V c = V b 2 V = V. Pont d s V less than pont c, so V d = V c V = 0 V. Ths s a consstency check makng one complete loop brngs us back to the potental at whch we started, namely 0 V. (b) The nformaton about the potentals s shown n the graph above. (c) Movng the ground to pont a doesn t change the basc analyss of part (a) or the potental dfferences found there. All that changes s that now V a = 0 V. Pont b s 2 V more than pont a because of the battery. So, V b = V a + 2 V = 2 V. Pont c s 2 V less than pont b, so V c = V b 2 V = 0 V. Pont d s V less than pont c, so V d = V c V = 4 V. Pont a s 4 V less than pont d, so V a = V d 4 V = 0 V. Ths brngs us back to where we started. The nformaton about the potentals s shown n the graph above Model: The voltage source/battery and the connectng wres are deal. Please refer to Fgure P Solve: Let us frst apply Krchhoff s loop law startng clockwse from the lower left corner: The output voltage s Vn +V n R (00 Ω) = 0 V = R + 00 Ω Vn Vout 00 Ω Vout = ( 00 Ω ) = ( 00 Ω) = R + 00 Ω V R+ 00 Ω For V 0 out = V n, the above uaton can be smplfed to obtan R: Vn 0 00 Ω = V R+ 00 Ω n R + 00 Ω = 000 Ω R = 900 Ω n 32.. Model: The battery and the connectng wres are deal. The fgure shows how to smplfy the crcut n Fgure P32. usng the laws of seres and parallel resstances. Havng reduced the crcut to a sngle uvalent resstance, we wll reverse the procedure and buld up the crcut usng the loop law and the juncton law to fnd the current and potental dfference of each resstor. Solve: From the last crcut n the dagram, = E 2 V 2 A Ω = Ω = Thus, the current through the battery s 2 A. As we rebuld the crcut, we note that seres resstors must have the same current and that parallel resstors must have the same potental dfference V. 4

5 n Step, the Ω resstor s returned to a 3 Ω and 3 Ω resstor n seres. Both resstors must have the same 2 A current as the Ω resstance. We then use Ohm s law to fnd V 3 = (2 A)(3 Ω) = V As a check, V + V = 2 V, whch was V of the Ω resstor. n Step 2, one of the two 3 Ω resstances s returned to the 4 Ω, 48 Ω, and Ω resstors n parallel. The three resstors must have the same V = V. From Ohm s law, V 4 = =.5 A 4 Ω V 48 = = A 48 Ω 8 V 3 = = A Ω 8 Resstor Potental dfference (V) Current (A) 3 Ω 4 Ω 48 Ω Ω Model: The connectng wres are deal. The capactors dscharge through the resstors. The fgure shows how to smplfy the crcut n Fgure P32.7 usng the laws of seres and parallel resstors and the laws of seres and parallel capactors. Solve: The 30 Ω and 20 Ω resstors are n parallel and are uvalent to a 2 Ω resstor. Ths 2 Ω resstor s n seres wth the 8 Ω resstor so the uvalent resstance of the crcut R = 20 Ω. The two 0 µf capactors are n seres producng an uvalent capactance of 30 µf. Ths 30 µf capactor s n parallel wth the 20 µf capactor so the uvalent capactance C of the crcut s 50 µf. The tme constant of ths crcut s τ = R C = (20 Ω)(50 µf) =.0 ms The current due to the three capactors through the 20 Ω uvalent resstor s the same as through the 8 Ω resstor. So, the voltage across the 8 Ω resstor follows the decay uaton V = V 0 e t/τ. For V = V 0 /2, we get t V 0 /2 = V 0 e t/.0 ms ln = 2.0 ms t = 0.9 ms 5