1.) How long must a 0.60-mm-diameter aluminum wire be to have a 0.52A current wh
ID: 1282560 • Letter: 1
Question
1.) How long must a 0.60-mm-diameter aluminum wire be to have a 0.52A current when connected to the terminals of a 1.5 V flashlight battery?
2.) What is the current if the wire is half this length?
3.) The current in an electric hair dryer is 10A
a.) How many electrons flow through the hair dryer in 5.0min ?
4.) A battery with an emf of 60 V is connected to the two capacitors shown in the figure(Figure 1) . Afterward, the charge on capacitor 2 is 360?C . (http://session.masteringphysics.com/problemAsset/1384664/2/30.P68.jpg)
a.)What is the capacitance of capacitor 2?
Explanation / Answer
1) resistivity of aluminium = 2.65*10-8
R= V/I =1.5/0.52 =2.8846
R=resistivity*length/area
length = R*area /resistivity =2.8846 * pi* (0.3*10-3)2 /( 2.65*10-8 ) =3077.741m
2) I1*L1 =I2*L2
L2 = .5*L1 =>I2 = 2*I1 =2*.52 =1.04A
3) q =10*5*60 =3000C
4)V1 = Q/C1 and V2 = Q/C2. Therefore: V = V1 + V2 = Q(1/C1 + 1/C2)
60 =360*10-6 *(1/C1 + 1/C2)
C2== 12uf
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