, and the number of electrons is 19. e e C. The negatively charged electrons move in the direction opposite to the conventional current flow.


 Ann Harvey
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1 Prolem 1. f current of 80.0 ma exists in metl wire, how mny electrons flow pst given cross section of the wire in 10.0 min? Sketch the directions of the current nd the electrons motion. Solution: The chrge tht moves pst the cross section is Δ Q = ( Δ t, nd the numer of electrons is ΔQ n = = e ( Δt e ( C s ( 10.0 min ( 60.0 smin = = C The negtively chrged electrons move in the direction opposite to the conventionl current flow electrons v Prolem. A smll sphere tht crries chrge q is whirled in circle t the end of n insulting string. The ngulr frequency of rottion is ω. Wht verge current does this rotting chrge represent? Solution: The period of revolution for the sphere is T = π ω, nd the verge q qω current represented y this revolving chrge is = = T π Prolem 3. n prticulr television picture tue, the mesured em current is 60.0 μa. How mny electrons strike the screen every second? Solution: Δ Q = ( Δ t nd the numer of electrons is ( Δt ( C s ( 1.00 s ΔQ n = 19 e e C electrons Prolem 4. A totl chrge of 6.0 mc psses through crosssectionl re of wire in.0 s. Wht is the current in the wire? 3 Δ Q C 3 Solution: The current is A= Δt.0 s 3.0 m A
2 Prolem 5. n the Bohr model of the hydrogen tom, n electron in the lowest energy stte moves t speed of m/s in circulr pth hving rdius of m. Wht is the effective current ssocited with this oriting electron? Solution: The period of the electron in its orit is T represented y the oriting electron is = πrv, nd the current ΔQ e ve Δt T π r 6 19 ( m s( C 11 π ( m C s = = = 1.05 ma Prolem 6. f kg of gold is deposited on the negtive electrode of n electrolytic cell in period of.78 h, wht is the current through the cell in this period? Assume tht the gold ions crry one elementry unit of positive chrge. Solution: The mss of single gold tom is m tom M 197 g m ol = N A tom s m ol g kg The numer of toms deposited, nd hence the numer of ions moving to the negtive electrode, is kg m n m kg tom Thus, the current in the cell is 1 19 ( ( C (.78 h( 3600 s1 h ΔQ ne = A = 159 ma Δt Δt
3 Prolem 7. A 00kmlong highvoltge trnsmission line.0 cm in dimeter crries stedy current of A. f the conductor is copper with free chrge density of electrons per cuic meter, how long (in yers does it tke one electron to trvel the full length of the cle? Solution: The drift speed of electrons in the line is v = d nqa = ne πd, or v d ( A = = ( m ( C π ( 0.00 m The time to trvel the length of the 00km line is then ( m s 3 L m 1 yr Δ t= = 4 v 7 = m s s d 7 yr Prolem 8.An luminum wire with crosssectionl re of 4.0 x 10 6 m crries current of 5.0 A. Find the drift speed of the electrons in the wire. The density of luminum is.7 g/cm 3. (Assume tht one electron is supplied y ech tom. Solution:Assuming tht, on verge, ech luminum tom contriutes one electron, the density of chrge crriers is the sme s the numer of toms per cuic meter. This is density ρ N A ρ n, msspertom M N A M or ( m ol (.7 g cm 3 ( 10 6 cm 3 1 m 3 n = = m 6.98 g m ol 8 3 The drift speed of the electrons in the wire is then 5.0 C s vd nea m s ( m 3 ( C( m Prolem 9. f the current crried y conductor is douled, wht hppens to the ( chrge crrier density nd ( electron drift velocity? Solution: ( The crrier density is determined y the physicl chrcteristics of the wire, not the current in the wire. Hence, n is unffected. ( The drift velocity of the electrons is vd = nqa. Thus, the drift velocity is douled when the current is douled.
4 Prolem 10. A lightul hs resistnce of 40 when operting t voltge of 10 V. Wht is the current through the lightul? 10 V Solution: A = 500 ma 40 Prolem 11. A person notices mild shock if the current long pth through the thum nd index finger exceeds 80 μa. Compre the mximum llowle voltge without shock cross the thum nd index finger with dryskin resistnce of nd wetskin resistnce of Solution: ( Δ = mx = ( A mx Thus, if V nd if = , ( Δ V mx = 3 V, =, ( Δ V mx = 0.16 V Prolem 1. Suppose tht you wish to fricte uniform wire out of 1.00 g of copper. f the wire is to hve resistnce of = 0.500, nd if ll of the copper is to e used, wht will e ( the length nd ( the dimeter of this wire? Solution: The volume of the copper is kg m V m 3 3 density kg m Since, V = A L, this gives A L= m. (1 ( From =, we find tht A 8 ρ m 8 A = L= L= ( m L nserting this expression for A into Eqution 1 gives ( m m 8 L = 7 3, which yields L = 1.8 m ( From eqution (1, 7 3 πd m A = =, or 4 L ( ( m m d= = πl π( 1.8 m 4 = m = 0.80 mm
5 Prolem 13. Clculte the dimeter of.0cm length of tungsten filment in smll lightul if its resistnce is πd Solution: From =, we otin A = =, or A 4 8 ( ( 4ρ L m.0 10 m 4 d m = 0.17 mm π π( Prolem 14. Eighteenguge wire hs dimeter of 1.04 mm. Clculte the resistnce of 15 m of 18guge copper wire t 0 C ( 10 m ( 15 m Solution: = A πd 4 π m ( Prolem 15.A potentil difference of 1 V is found to produce current of 0.40 A in 3.m length of wire with uniform rdius of 0.40 cm. Wht is ( the resistnce of the wire nd ( the resistivity of the wire? 1 V Solution: ( A ( From, =, A ( 30 ( m A π ρ = = = L 3. m m Prolem 16. A length L 0 of copper wire hs resistnce 0. The wire is cut into three pieces of equl length. The pieces re then connected s prllel lengths etween points A nd B. Wht resistnce will this new wire of length L 0 /3 hve etween points A nd B? Solution: The new wire hs length L L0 3 A = 3A. Thus, its = nd crosssection 0 resistnce is ρ( L = A 3A 9 A Prolem 17. A wire 50.0 m long nd.00 mm in dimeter is connected to source with potentil difference of 9.11 V, nd the current is found to e 36.0 A. identify the metl of the wire V Solution: The resistnce is 0.53, so the resistivity of the 36.0 A metl is 3 ( πd 4 ( 0.53 π( m A ρ = L L m Thus, the metl is seen to e silver. ( m
6 Prolem 18. A 4.0 resistor, n 8.0 resistor, nd 1 resistor re connected in series with 4V ttery. Wht re ( the equivlent resistnce nd ( the current in ech resistor? (c epet for the cse in which ll three resistors re connected in prllel cross the ttery. Solution: ( eq = + + =+ + + = 4 ( The sme current exists in ll resistors in series comintion. 4 V 4 eq 1.0 A (c f the three resistors were connected in prllel, eq = + + = + + = esistors in prllel hve the sme potentil difference cross them, so 4 V A, 4 V 8 = = A, nd 4 V 1 = = A Prolem 19. A 9.0 resistor nd 6.0 resistor re connected in series with power supply. ( The voltge drop cross the 6.0 resistor is mesured to e 1 V. Find the voltge output of the power supply. ( The two resistors re connected in prllel cross power supply, nd the current through the 9.0 resistor is found to e 0.5 A. Find the voltge setting of the power supply. Solution: ( The current through this series comintion is c c 1 V.0 A 6.0 c Therefore, the terminl potentil difference of the power supply is Δ V = eq = (.0 A( = 30 V ( When connected in prllel, the potentil difference cross either resistor is the voltge setting of the power supply. Thus, ( ( Δ V = 99 = 0.5 A 9.0 =.3 V
7 Prolem 0. ( Find the equivlent resistnce etween points nd in the following figure ( Clculte the current in ech resistor if potentil difference of 34.0 V is pplied etween points nd. Solution: ( The equivlent resistnce of the two prllel resistors is p = + = Thus, ( = 4+ p + 9 = = 17.1 ( 34.0 V 1.99 A 17.1, so 4 = 9 = 1.99 A Also, ( V ( ( Δ = = 1.99 A 4.1 = 8.18 V p p Then, nd p 8.18 V p 8.18 V A A
8 Prolem 1. Find the equivlent resistnce of the circuit in figure shown. Solution: The equivlent resistnce of the prllel comintion of three resistors is p = + + = Hence, the equivlent resistnce of the circuit connected to the 30 V source is eq = 1 + p = = 15 Prolem. Find the equivlent resistnce of the circuit in the figure shown Solution: The rules for comining resistors in series nd prllel re used to reduce the circuit to n equivlent resistnce in the stges shown elow. The result is = 9.8. eq
9 Prolem 3. Consider the circuit shown in the following figure. Find ( the current in the 0.0 resistor nd ( the potentil difference etween points nd. Solution: Turn the circuit given in Figure 90 counterclockwise to oserve tht it is equivlent to tht shown in Figure 1 elow. This reduces, in stges, s shown in the following figures. 5.0 V c V Figure 1 Figure 5.0 V V 1.9 Figure 3 Figure 4 From Figure 4, 5.0 V 1.93 A 1.9 ( From Figure 3, ( Δ V = = ( 1.93 A(.94 = 5.68 V ( From Figures 1 nd, the current through the 0.0 resistor is 5.68 V A c
10 Prolem 4. Two resistors, A nd B, re connected in prllel cross 6.0V ttery. The current through B is found to e.0 A. When the two resistors re connected in series to the 6.0V ttery, voltmeter connected cross resistor A mesures voltge of 4.0 V. Find the resistnces of A nd B. Solution: First, consider the prllel cse. The resistnce of resistor B is B 6.0 V B A B n the series comintion, the potentil difference cross B is given y ( Δ V = ( Δ V = 6.0 V4.0 V =.0 V B ttery A The current through the series comintion is then B.0 V A s B A nd the resistnce of resistor A is A A, ( 4.0 V s Prolem 5. The resistnce etween terminls nd in the following figure is 75. f the resistors leled hve the sme vlue, determine. Solution: The equivlent resistnce is eq = + p, where p is the totl resistnce of the three prllel rnches; 1 1 ( 30 ( p = + + = + = Thus, ( ( ( = + = , which reduces to ( = 0 or ( ( = 0. Only the positive solution is physiclly cceptle, so = 55
11 Prolem 6. Find the current in the 1 resistor in the following figure Solution: The resistors in the circuit cn e comined in the stges shown elow = to yield n equivlent resistnce of d ( c e d c e d Figure 1 18 V 18 V Figure d d d V Figure 3 ( 18 V Figure 4 d 18 V From Figure 5, 3.14 A d (. 18 V Figure 5 Δ = = 3.14 A 3011 = 8.57 V. Then, from Figure 4, ( V d ( ( Now, look t Figure nd oserve tht d d 8.57 V 1.71 A , so ( V ( ( Δ = e = 1.71 A 3.0 = 5.14 V e Finlly, from Figure 1, e 5.14 V A
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