Chapter E  Problems


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1 Chpter E  Prolems Blinn College  Physics Terry Honn Prolem E.1 A wire with dimeter d feeds current to cpcitor. The chrge on the cpcitor vries with time s QHtL = Q 0 sin w t. Wht re the current nd current density in the wire s functions of time? Solution to E.1 The current is the the time derivtive of the chrge. The current density is just the current per re. I = Q t = Q 0 w cos w t J = I A = Q 0 w cos w t p d 2 ë4 Prolem E.2 The current s function of time vries s IHtL = 15 sinh120 p tl in SI units. Wht is the totl chrge tht flows etween 0 nd 1 s. (This is hlf cycle of stndrd 60 Hz AC frequency.) 120 Solution to E.2 Q = Ÿ IHtL t = Ÿ 0 1ê sinh120 p tl t = p cosh120 p tl 0 1ê120 = 1 4 p C = C Prolem E.3 The dimeter of stndrd 12 guge wire is 2.05 mm. Suppose in stndrd houshold wiring 15 A current flows through 20 m length of copper wire. () Wht is the resistnce in the wire nd wht is the voltge drop cross the wire? () For ech copper tom there is one free electron ville for conduction. The density (mss/volume) of copper is 8920 kgëm 3 nd its tomic mss is u. Use the conversion 6.02µ10 26 u = 1 kg to find the mss of ech tom. From tht find the numer of toms per volume; this for copper is the sme s the numer of chrge crriers per volume. Clculte the drift velocity of the free electrons nd find the totl time it tkes one electron to migrte the full length of the wire. Solution to E.3 () The resistivity of copper is 1.7µ108 Wÿm. The crosssectionl re of the wire is The voltge drop cross wire is given y Ohm's lw. A = p H ê2L 2 = µ106 m 2 R = r { A = 1.7µ108 µ20 = = W A
2 2 Chpter E  Prolems V = I R = 15µ = 1.55 V () I = n q v d A q is the chrge of the chrge crriers, for metl q = e. A is the crosssectionl re, found ove. v d is the drift velocity nd n is the numer of chrge crriers per volume. The mss of ech tom is: 1 kg m tom = uµ 6.02µ10 26 = u µ1025 kg Since there is one electron per tom of copper tht contriutes to conduction, n is the sme s the numer of toms per volume. Since the density is the mss per volume we cn write: n = rêm tom. n = free electrons volume = I = n q v d A ï v d = free electrons µ toms tom volume = 1µ The time to trvel the length of the wire is given y { = v d t r = m tom 8920 kgëm µ1028 = µ1025 m 3 I n e A = 15 n 1.60µ1019 = A µ104 = 3.36µ104 m s t = { v d = 20 v d = s = 16.5 hr Note tht the electricl signl trvels much fster thn tht; tht speed is on the order of the speed of light. Prolem E.4 Suppose wire with restnce R is stretched y 30% keeping the volume of the wire fixed. Wht is the new resistnce of the wire? Solution to E.4 If wire is stretched its volume stys constnt so tht A { = A {. Thus { = 1.30 { nd Since R = r { ê A we get the rtio: A A = { A = 1.30 ï { A = R R = { ê{ A êa = ê1.30 = = 1.69 ï R = 1.69 R Prolem E.5 When the temperture of wire is decresed y 10 C its resistnce decreses y 6 %. Wht is its temperture coefficient? Solution to E.5 D R = R 0 ï D R R 0 = D T ï = H10L ï = 0.006êC Prolem E.6 The precise definition of the temperture coefficient is = 1 r Show tht if is constnt we get: r = r 0 e HTT 0 L. Use the pproximtion tht for smll x (x ` 1), e 1 + x, to show tht for smll D T = T  T 0 : r T. r = r 0 e H r + HT  T 0 LD.
3 Chpter E  Prolems 3 Solution to E.6 = 1 r Integrting this nd using r = r 0 t T = T 0 gives r T = ln r T ï ln r = T Ÿ r0 r ln r = Ÿ T0 T T ï ln r  ln r0 = HT  T 0 L. Exponentiting this gives: r = r 0 e HTT 0 L. If D T = T  T 0 is smll then x = HT  T 0 L is smll nd we cn sue the expnsion. r = r 0 e HTT 0 r + HT  T 0 LD Prolem E.7 A hir dryer desiigned for stndrd US outlet hs power rting of 1500 W. () How much current does it drw? () Wht is its resistnce? Solution to E.7 For stndrd outlet we hve V rms = 120 V. This is clled the rootmensqured voltge. We will discuss this in detil in the AC circuit chpter. For now we just need to know tht for purely resistive AC circuit we cn tret it like DC if we use the rms voltge nd current nd the verge power. () P = V I ï I = P V = = 12.5 A V = 120 V nd P = 1500 W () P = V2 R ï R = V2 P = = 9.6 W Prolem E.8 Suppose purely resistive Ohmic device designed for stndrd outlet (V = 120 V) is operted t too low voltge, 105 V. By wht percent would the power of the device e decresed? Solution to E.8 The power decreses y 23.4 %. P = V2 R ï P P = J V V N2 = J N2 = Prolem E.9 kwÿhr is the unit of energy used in electric ills. Wht is 1 kwÿhr in J? If the electric utility chrges $0.09 per kwÿhr, then wht is the cost of running 100 W ul constntly for 31 dy month? Solution to E.9
4 4 Chpter E  Prolems kwÿh = 1000 W µ3600 s = 3.6µ10 6 J Do the clcultion of cost using just kwÿh insted of J. For 100 W ul we hve Power = 0.10 kw. 24 hr Energy = Powerµtime = 0.10 kwµ31 dysµ = 74.4 kwÿhr dy $0.09 Cost = EnergyµRte = 74.4 kwÿhrµ = $6.70 kwÿhr Prolem E.10 A student tries to determine the internl resistnce of ttery. A voltmeter connected cross the ttery without lod reds the EMF E. When the ttery is connected cross known lod resistnce R 0, the voltmeter reds cross the ttery reds V 0. Wht is the internl resistnce r? Solution to E.10 V t = E  I r The voltmeter cross the ttery reds it's terminl voltge V t = V 0. This is cross the resistnce R 0 giving the current. V 0 = I R 0 ï I = V 0 R 0 V t = E  I r ï V 0 = E  V 0 R 0 r ï r = R 0 E V 01 Prolem E.11 An unknown resistnce R is connected cross fixed voltge source. When 200 W resistor is plced in prllel with it the current delivered y the source qudruples. Wht is the resistnce? Solution to E.11 Qudrupling the current t fixed voltge implies 1/4 the resistnce. R eq = R 4 = J 1 R N1 ï 4 R = 1 R ï R = 600 W Prolem E.12 12W 10W 15W 90V 20W Complete the tle elow with the voltge cross, the current through nd the totl power dissipted in ech resistor. V I 12 W 10 W 15 W 20 W
5 Chpter E  Prolems 5 P Also, give the totl current nd power delivered y the ttery. How re these vlues relted to those in the tle? Solution to E.12 Once one element in column is known the other two cn esily e found using V = I R nd P = V I. The 20 W resistor is connected cross the 90 V source. V 20 = 90 ï I 20 = V 20 = 90 R = 4.5 ï P 20 = V 20 I 20 = 90µ4.5 = 405 The equivlent resistnce of the other three resistors is: 12 + J N1 = 18. All of the current through this equivlent flows through the 12 W resistor. I 12 = I 18 = = 5 ï V 12 = I 12 R 12 = 5µ12 = 60 ï P 12 = V 12 I 12 = 60µ5 = 300 The voltges cross the other two prllel resistors re equl V 10 = V 15 = 90  V 12 = = 30 ï I 10 = V 10 R 10 = = 3 nd I 15 = V 15 R 15 = = 2 ï P 10 = V 10 I 10 = 30µ3 = 90 nd P 15 = V 15 I 15 = 30µ2 = W 10 W 15 W 20 W V 60 V 30 V 30 V 90 V I 5 A 3 A 2 A 4.5 A P 300 W 90 W 60 W 405 W The overll equivlent is: R eq = J N1 = 180 = The totl current delivered y the ttery is 19 I tot = V tot = 90 = 9.5 A nd the totl power delivered y the ttery is R eq 180ê19 P tot = V tot I tot = 90µ9.5 = 855 W. Note tht the totl current is the sum of the currents through the two rnches 9.5 = nd the totl power is the sum of ll powers 855 = Prolem E.13 Suppose resistnce R is put on ech edge of cue. Wht is the equivlent resistnce etween opposite corners of the cue? Solution to E.13 c d c d c Using symmetry we cn identify corners s hving the sme voltge. Nodes lelled y c re one resistor wy from node nd must hve the sme voltge. Similrly nodes lelled y d should hve the sme voltge. We my then, since they re d
6 6 Chpter E  Prolems nd must hve the sme voltge. Similrly nodes lelled y d should hve the sme voltge. We my then, since they re t the sme voltge, short out the c nodes nd short out the d nodes. There re three resistors etween nd c; these re in prllel nd thus hve n equivlent of Rê3. There re six resistors etween c nd d, which thus hve n equivlent of Rê6. The threee etween d nd give Rê3. The resulting three resistors re in series giving n overll equivlent of: R eq = R 3 + R 6 + R 3 = 5 6 R Prolem E.14 Wht is the equivlent resistnce etween nd? 12W 8W 16W 12W 24W 96W Solution to E.14 12W 3 16W 24W 48W 1 72W 96W There re no resistors in this network tht re either in series or prllel, so we must use the node reduction formul. Applying this to the three internl resistors we get: R = J N1 = 4 R 12 = R 1 R 2 = 12µ24 = 72 R 4 R 13 = R 1 R 3 = 12µ8 = 24 R 4 2 R 23 = R 2 R 3 = 24µ8 = 48 R 4 Now identify tht the 24 W nd 12 W resistors re in prllel s re the 48 W nd 16 W resistors re in series giving: The 20 W, 72 W nd 96 W resistors re then in prllel: J N1 + J N1 = = 20. resistors. The resulting two
7 Chpter E  Prolems 7 R eq = J N1 = W. Prolem E.15 Wht is the equivlent resistnce etween nodes nd? 1W 1W 3W 1W 5W Solution to E.15 This is n exmple where no two resistors re in series or prllel so we must use the nodereduction method. The three 1 W resistors re the esiest to eliminte (insted of the 1 W, 3 W nd 5 W ) ecuse the new resistors re equl. R = = R 1 R 2 R 3 3 ï R ij = R i R j = 1µ1 R 1ê3 = 3 1 3W 2 3W 3W 3W 3 5W Comine the pir of 3 W resistors in the lower left in prllel nd lso comine the 3 W nd 5 W resistors in the lower right in prllel. The resulting two resistors re then in series nd this is then in prllel with the 3 W t the top. J N1 + J N1 = 27 8 ï R eq = K ê8 O1 = W. Prolem E.16 4mF 9V 6mF 12mF 18mF () Wht is the equivlent cpcitnce cross the 9 V source? () Complete the tle elow with the voltge cross ech cpcitor nd the chrge on ech. V Q 18 mf 6 mf 4 mf 12 mf Solution to E.16
8 8 Chpter E  Prolems () The 4 mf nd 12 mf cpcitors re in series, so their equivlent is: J N1 = 3 mf. This is in prllel with the 6 mf cpcitor, giving = 9 mf. This, in turn, is in series with the 18 mf giving C eq = J N1 = 6 mf. () The chrge on the 18 mf is the sme s on the equivlent. Q 18 = Q eq = C eq V tot = 6µ9 = 54 mc. This is the chrge delivered y the ttery. The voltge cross it is V 18 = Q 18 C 18 = = 3 V. The voltge cross the 6 mf cpcitor nd the equivlent 3 mf cpcitor is: V 3 = V 6 = V tot  V 18 = 93 = 6 V. The chrge on the 6 mf cpcitor is: Q 6 = C 6 V 6 = 6µ6 = 36 mc The chrges on the 4 mf nd 12 mf cpcitors re equl to tht on their equivlent: Q 4 = Q 12 = Q 3 = C eq V 3 = 3µ6 = 18 mc. The voltges cross the 4 mf nd 12 mf cpcitors cn now e found: V 4 = Q 4 C 4 = 18 4 = 4.5 V nd V 12 = Q 12 C 12 = = 1.5 V 18 mf 6 mf 4 mf 12 mf V 3 V 6 V 4.5 V 1.5 V Q 54 mc 36 mc 18 mc 18 mc Prolem E.17 Wht is the equivlent cpcitnce etween nd? 2mF 8mF 3mF 3mF 9mF 6mF Solution to E.17 A node is point of constnt voltge in circuit. The two wires without components re nodes. To simplify this circuit shrink these wires to point. It follows, then, tht the 2 mf nd 8 mf cpcitors re in prllel (giving = 10), s re the 3 mf, 3 mf nd 9 mf (giving = 15). The two equivlents re then in series J N1 = 6 nd this is in prllel with the 6 mf cpcitor. The overll equivlent is then: C eq = = 12 mf. Prolem E.18 Give set of 5 liner equtions tht cn e solved for the currents. You need not solve the equtions.
9 Chpter E  Prolems 9 2W I 4 2V 3W 4W 9W I 3 6V I 5 I 2 I 1 8V 7V 7W 5W 5V Solution to E.18 The junction rule gives: I 1 = I 2 +I 3 + I 5 I 3 + I 5 = I 4 I 2 + I 4 = I 1 Recll tht ny one of the junction rule equtions is the sum of the others, so to get n independent set eliminte one of these three equtions. The loop rule gives: 79 I 24 I 1 +8 = I 4 +9 I I 3 +6 = I 35 I I 5 = 0
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