V bulb = V/N = (110 V)/8 = 13.8 V. We find the resistance of each bulb from 2 bulb bulb 7.0 W = (13.8 V) 2 /R bulb, which gives R bulb = 27 Ω.

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1 HPTE iruits 4. We fin the internl resistne from V = r; 9.8 V = [.0 V (60 )r], whih gives r = Ω. euse the terminl voltge is the voltge ross the strter, we hve V = ; 9.8 V = (60 ), whih gives = 0.6 Ω. 5. When the uls re onnete in series, the equivlent resistne is series = i = 4 ul = 4(90 Ω) = 360 Ω. When the uls re onnete in prllel, we fin the equivlent resistne from / prllel = (/ i ) = 4/ ul = 4/(90 Ω), whih gives prllel = 3 Ω. 7. f we use them s single resistors, we hve = 5 Ω; = 70 Ω. When the resistors re onnete in series, the equivlent resistne is series = i = = 5 Ω 70 Ω = 95 Ω. When the resistors re onnete in prllel, we fin the equivlent resistne from / prllel = (/ i ) = (/ ) (/ ) = [/(5 Ω)] [/(70 Ω)], whih gives prllel = 8 Ω.. We n reue the iruit to single loop y suessively 7 omining prllel n series omintions. We omine n, whih re in series: 7 = =.8 kω.8 kω = 5.6 kω. We omine n 7, whih re in prllel: / 8 = (/ ) (/ 7 ) = [/(.8 kω)] [/(5.6 kω)], whih gives 8 =.87 kω. We omine n 8, whih re in series: 9 = 8 =.8 kω.87 kω = 4.67 kω. We omine n 9, whih re in prllel: 9 /0 = (/ ) (/ 9 ) = [/(.8 kω)] [/(4.67 kω)], whih gives 0 =.75 kω. We omine 0 n, whih re in series: eq = 0 =.75 kω.8 kω = 4.6 kω. 8 0 eq 4. n series the urrent must e the sme for ll uls. f ll uls hve the sme resistne, they will hve the sme voltge: V ul = V/N = (0 V)/8 = 3.8 V. We fin the resistne of eh ul from P ul = V / ; ul ul 7.0 W = (3.8 V) / ul, whih gives ul = 7 Ω. 6. The equivlent resistne of the two resistors onnete in series is s =. We fin the equivlent resistne of the two resistors onnete in prllel from / p = (/ ) (/ ), or p = /( ). The power issipte in resistor is P = V /, so the rtio of the two powers is P p /P s = s / p = ( ) / = 4. When we expn the squre, we get = 4, or = ( ) = 0, whih gives = =.6 kω.

2 eq From ove, we hve V =.85 V. 8. We n reue the iruit to single loop y suessively omining prllel n series omintions. We omine n, whih re in series: 7 = =.0 kω.0 kω = 4.40 kω. We omine n 7, whih re in prllel: / 8 = (/ ) (/ 7 ) = [/(.0 kω)] [/(4.40 kω)], whih gives 8 =.47 kω. We omine n 8, whih re in series: 9 = 8 =.0 kω.47 kω = 3.67 kω. We omine n 9, whih re in prllel: /0 = (/ ) (/ 9 ) = [/(.0 kω)] [/(3.67 kω)], whih gives 0 =.38 kω. We omine 0 n, whih re in series: eq = 0 =.38 kω.0 kω = 3.58 kω. The urrent in the single loop is the urrent through : 6 = = V/ eq = ( V)/(3.58 kω) = 3.36 m. For V we hve V = 0 = (3.36 m)(.38 kω) = 4.63 V. This llows us to fin 5 n 4 ; 5 = V / = (4.63 V)/(.0 kω) =. m; 4 = V / 9 = (4.63 V)/(3.67 kω) =.6 m. For V we hve V = 4 8 = (.6 m)(.47 kω) =.85 V. This llows us to fin 3,, n ; 3 = V / = (.85 V)/(.0 kω) = 0.84 m; = = V / 7 = (.85 V)/(4.40 kω) = 0.4 m. 9. () When the swith is lose the ition of to the prllel set will erese the equivlent resistne, so the urrent from the ttery will inrese. This uses n inrese in the voltge ross, n orresponing erese ross n. The voltge ross inreses from zero. Thus we hve V n V inrese; V 3 n V 4 erese. () The urrent through hs inrese. This urrent is now split into three, so urrents through n erese. Thus we hve (= ) n inrese; 3 n 4 erese. () The urrent through the ttery hs inrese, so the power output of the ttery inreses.

3 () efore the swith is lose, = 0. We fin the resistne for n in prllel from / = (/ i ) = / = /(00 Ω), whih gives = 50 Ω. For the single loop, we hve = = V/( ) = (45.0 V)/(00 Ω 50 Ω) = This urrent will split evenly through n : 3 = 4 =! =!(0.300 ) = fter the swith is lose, we fin the resistne for,, n in prllel from / = (/ i ) = 3/ = 3/(00 Ω), whih gives = 33.3 Ω. For the single loop, we hve = = V/( ) = (45.0 V)/(00 Ω 33.3 Ω) = This urrent will split evenly through,, n : = 3 = 4 = ⅓ = ⅓(0.338 ) = 0.3. S V V () When the swith is opene, the removl of resistor from the prllel set will inrese the equivlent resistne, so the urrent from the ttery will erese. This uses erese in the voltge ross, n orresponing inrese ross. The voltge ross ereses to zero. r 3 S r Thus we hve V n V 3 erese; V inreses. () The urrent through hs erese. The urrent through hs inrese. The urrent through hs erese to zero. Thus we hve (= ) n 3 erese; inreses. () euse the urrent through the ttery ereses, the r term ereses, so the terminl voltge of the ttery will inrese. () When the swith is lose, we fin the resistne for n in prllel from / = (/ i ) = / = /(5.50 Ω), whih gives =.75 Ω. For the single loop, we hve = V/( r) = (.0 V)/(5.50 Ω.75 Ω 0.50 Ω) =.37. For the terminl voltge of the ttery, we hve V = r =.0 V (.37 )(0.50 Ω) =.3 V. (e) When the swith is open, for the single loop,

4 we hve = V/( r) = (.0 V)/(5.50 Ω 5.50 Ω 0.50 Ω) = When we inlue the urrent through the ttery, we hve six unknowns. For the onservtion of urrent, we hve juntion : = ; juntion : = 3 5 ; juntion : 5 = 4. For the three loops inite on the igrm, we hve loop : 5 = 0; (0 Ω) 5 (0 Ω) (5 Ω) = 0; loop : = 0; 3 ( Ω) 4 ( Ω) 5 (0 Ω) = 0; loop 3: 4 = V (5 Ω) 4 ( Ω) = When we solve these six equtions, we get = 0.74, = 0., 3 = 0.66, 4 = 0.9, 5 = 0.007, = We hve rrie n extr eiml ple to show the greement with the juntion equtions. 33. The given urrent is = For the onservtion of urrent t point, we hve 3 =, or =. For the top loop inite on the igrm, we hve loop : r r = 0;.0 V (.0 Ω) (8.0 Ω).0 V ( 0.30 )(.0 Ω) ( 0.30 )(0 Ω) ( Ω) = 0, whih gives =.30. Thus we hve 3 = =.30 ( 0.30 ) =.60. For the ottom loop inite on the igrm, we hve loop : 3 r 3 r 3 = 0; (.60 )(.0 Ω) (.60 )(8 Ω) ( 0.30 )(0 Ω).0 V ( 0.30 )(.0 Ω) (.60 )(5 Ω) = 0, whih gives = 70 V. e r r g 3 r f For the onservtion of urrent t point, we hve 3 =. For the two loops inite on the igrm, we hve loop : r r = 0;.0 V (.0 Ω) (8.0 Ω).0 V (.0 Ω) (0 Ω) ( Ω) = 0; loop : 3 3 r 3 3 r 3 = 0; 6.0 V 3 (.0 Ω) 3 (8 Ω) (0 Ω).0 V (.0 Ω) 3 (5 Ω) = 0. When we solve these equtions, we get = 0.77, = 0.7, 3 = For the terminl voltge of the 6.0-V ttery, we hve V fe = 3 3 r 3 = 6.0 V (0.055 )(.0 Ω) = 5.95 V. e r r g 3 3 r 3 f

5 4. () We fin the pitne from τ = ; s = (5 0 3 Ω), whih gives = F = 3.7 nf. () The voltge ross the pitne will inrese to the finl stey stte vlue. The voltge ross the resistor will strt t the ttery voltge n erese exponentilly: V = e t/τ ; 6.0 V = (4.0 V)e t/(55 μs), or t/(55 μs) = ln(4.0 V/6.0 V) = 0.405, whih gives t = μs. 43. The hrge on the pitor inreses with time to finl hrge Q 0 : Q = Q 0 ( e t/τ ). When we express the store energy in terms of hrge we hve U = ½V = ½Q / = ½(Q /)( e t/τ 0 ) = U mx ( e t/τ ). We fin the time to reh hlf the mximum from ½ = ( e t/τ ), or e t/τ = /v, whih gives t =.3τ. S 45. () For the onservtion of urrent t point, we hve = 3. For the loop on the right, we hve Q/ = 0, or = Q/. For the outsie loop, we hve Q/ = 0, or = Q/. The urrent 3 is hrging the pitor: 3 = Q/t. When we use these results in the juntion eqution, we get ( Q/)/ = (Q/ ) Q/t, whih eomes = Q/t ( )Q/. This hs the sme form s the simple iruit euse we hve no simple series or prllel onnetions, we nlyze the iruit. On the igrm, we show the potentil ifferene pplie etween points n, n the four urrents. For the onservtion of urrent t points n, we hve V 3 = 3 = 4. () The urrent 3 is hrging the pitor : 3 = Q 3 /t. The urrent 4 is hrging the pitor : 4 = Q 4 /t. For W loop, we hve Q 3 / = 0, or Q 3 / = ( 3 ) = ( Q 3 /t). () For W loop, we hve Q 4 / = 0, or Q 4 / = ( 4 ) = ( Q 4 /t). (3) For the pth, we hve V = (Q 3 / ) (Q 4 / ). (4) f we ifferentite this, we get V /t = 0 = (/ )(Q 3 /t) (/ )(Q 4 /t), or Q 4 /t = ( / ) Q 3 /t. (5) When we omine (3) n (5), we get Q 4 / = [ ( / ) Q 3 /t]. (6) When we omine () n (6), we get Q 4 / = ( / )Q 3 [ ( )/ ] Q 3 /t. (7) When we omine (4) n (7), we get V = [( )/ ]Q 3 [ ( )/ ] Q 3 /t. V 4

6 This hs the sme form s the simple iruit: = Q/t Q/, with = ( )/, n = /( ). Thus the time onstnt is τ = ( )/( ) = (8.8 Ω)(4.4 Ω)(0.48 μf 0.4 μf)/(8.8 Ω 4.4 Ω) =. μs.