The automobile starter motor with an equivalent resistance of 0.0500 2 connected to a 12.0-V battery with a 0.0100-22 internal resistance draws a current of 200 A, has 10.0 V applied to it, and receives 2000 W of power. Additional resistance from corrosion can significantly affect these values.
Calculations Involving Electric Circuits
To solve the problems related to the automobile starter motor and its electrical properties, we need to apply principles from electric circuit theory including Ohm's law and calculations involving power.
a) Current to the Motor
When a 12.0-V battery with a 0.0100-2¢ internal resistance is connected to a motor with an equivalent resistance of 0.0500 2, the total resistance in the circuit is the sum of both resistances, which is 0.0600 2. Applying Ohm's law (V = IR), where V is the voltage, I is the current, and R is the resistance, we find the current is I = V/R = 12.0 V / 0.0600 2 = 200 A.
b) Voltage Applied to the Motor
The voltage applied to the motor is the voltage across the motor's resistance, which is the total voltage minus the voltage drop across the battery's internal resistance. The voltage drop across the internal resistance (0.0100 2) is I * r = 200 A * 0.0100 2 = 2.0 V. Therefore, the voltage applied to the motor is 12.0 V - 2.0 V = 10.0 V.
c) Power Supplied to the Motor
The power supplied to the motor is calculated using P = IV, where P is power, I is current, and V is the voltage applied to the motor. P = 200 A * 10.0 V = 2000 W (or 2 kW).
d) Calculations with Additional Resistance
When the battery connections are corroded, an additional resistance of 0.0900 2 is added to the circuit. The total resistance now is 0.0500 2 (motor) + 0.0100 2 (internal) + 0.0900 2 (additional) = 0.1500 2. Repeating the calculations, the current to the motor is now I = V/R = 12.0 V / 0.1500 2 = 80 A. The voltage applied to the motor will also change as well as the power supplied, which will both decrease due to the increased resistance.