The starting motor of an automobile is turning too slowly, and the mechanic has to decide whether to replace the motor, the cable, or the battery. The manufacturer's manual says that the 12V battery should have no more than 0.02ohm internal resistance, the motor no more than 0.2ohm resistance, and the cable no more than 0.04ohm resistance. The mechanic turns on the motor and measures 11.4V across the battery, 3.0V across the cable, and a current of 50A. which part is defective?
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From Ohm's law .. internal resistance, r = voltage 'drop' across component / current
For cable .. r = 3.0V / 50A = 0.06Ω .. greater than 0.04Ω, ►indicating a problem
For the battery .. r = (12.0 - 11.4)V loss / 50A = 0.012Ω .. less than 0.02Ω .. OK
For the motor .. r = (11.4 - 3.0)V / 50A = 0.168Ω .. less than 0.20Ω .. OK
For cable .. r = 3.0V / 50A = 0.06Ω .. greater than 0.04Ω, ►indicating a problem
For the battery .. r = (12.0 - 11.4)V loss / 50A = 0.012Ω .. less than 0.02Ω .. OK
For the motor .. r = (11.4 - 3.0)V / 50A = 0.168Ω .. less than 0.20Ω .. OK
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Total resistance = R(int.) + R(motor) + R(cable)
R(Total) = .02 + .2 + .04 = 0.26 ohm
V = i R (total)
i = V/R(total) = 12V/0.26 ohm = 46.15 Amp
Voltage drop across cable should be: V(cable) = 46.15 * 0.04 = 1.846 V
but the actual voltage drop was 3.0 V which is more than expected. This indicated that the resistance in cable was greater than 0.04 ohm cause higher voltage drop.
The Cable was the problem.
R(Total) = .02 + .2 + .04 = 0.26 ohm
V = i R (total)
i = V/R(total) = 12V/0.26 ohm = 46.15 Amp
Voltage drop across cable should be: V(cable) = 46.15 * 0.04 = 1.846 V
but the actual voltage drop was 3.0 V which is more than expected. This indicated that the resistance in cable was greater than 0.04 ohm cause higher voltage drop.
The Cable was the problem.