An automobile starter motor has an equivalent resistance of 0.0510 ω and is supplied by a 12.0 v battery with a 0.0090 ω internal resistance. (a) what is the current (in a) to the motor? a (b) what voltage (in v) is applied to it? v (c) what power (in kw) is supplied to the motor? kw (d) repeat these calculations for when the battery connections are corroded and add 0.0900 ω to the circuit. (significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.) current a voltage v power kw

a) 200A

b) 10.2V

c) 2.04kW

d)

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

Where R is the equivalent resistance of the resistors in series

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

The power dissipated supplied to the motor is given by:

now solving adding a 0.0900 ohm resistor:

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form