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phy lab 8

# phy lab 8 - 8.1.2 Batteries in Series I 1 V=E-I r r =...

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8.1.2 Batteries in Series I 1. V=E-Ir r = {(V-E)/I} r = (4.25V-4.44V)/0.1A = 1.9 2. E’ = E1+E2+E3 = 1.479V+1.479V+1.472V = 4.43V 3. Percent Difference = {abs(E-E’)/E’}x100% {abs(4.44-4.43)/4.43}x100% = 0.23% 4. Internal resistance combined (r) = 1.9 Individual battery resistance (1.9/3) = 0.63 0.4753 was what was found in 8.1.1, which differs by .15 which seems to be consistent on how the resistance of the batteries also add up. 8.1.4 Exercise 1. E’-E+r’I+rI=0 so I = (E-E’)/(r’+r) 2. 3. I = 0.5V/(2 x 0.4753 ) = 0.52594A Power = (I^2 R) = (0.52594^2)(0.4753 ) = 0.131496 J/s Heat in one hour = (0.131496 J/s)(60s) = 7.89 J 4. lkdfjg 5. A fuse contains a strip of composed of different metals that are

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phy lab 8 - 8.1.2 Batteries in Series I 1 V=E-I r r =...

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