Question

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

Answer 1

a) 200A

b) 10.2V

c) 2.04kW

d)

I=80A

V=4.08V

P=0.326kW

Step-by-step 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:

`I=(V)/(R)`

Where R is the equivalent resistance of the resistors in series

`R=0.0510+0.0090=0.0600[ohm]`

`I=(12.0)/(0.0600)=200A`

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

`V_m=V*(R_m)/(R_m+R_s)\\V_m=12.0*(0.0510)/(0.0600)\\V_m=10.2V`

The power dissipated supplied to the motor is given by:

`P=I^2*R_m\\P=(200)^2*0.0510=2.04kW`

now solving adding a 0.0900 ohm resistor:

`I=(12.0)/(0.15)=80A`

`V_m=12.0*(0.0510)/(0.15)\\V_m=4.08V`

`P=I^2*R_m\\P=(200)^2*0.0510=0.326kW`