Two 1.50-V batteries
Two 1.50-V batteries
Homework Answers
a) If R is the resistance of the bulb we can use R =
V/I
total resistance of the
circuit=0.21+0.16+R=0.37+R
total voltage=2*1.5=3v
current=3/0.37+R
0.37+R=3/600*10^-3=5
R=5-0.37=4.63 ohm
b) total power = V*I = 0.6*3 = 1.8W
Power in the internal resistance = I^2*R = 0.6^2 * 0.37 =
0.1332W
Therefore fraction of total power dissipated in the batteries =
0.1332/ 1.8 = 0.074
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