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

Answer:

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:

$I=\frac{V}{R}$

Where R is the equivalent resistance of the resistors in series

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

$I=\frac{12.0}{0.0600}=200A$

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

$V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V$

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:

$I=\frac{12.0}{0.15}=80A$

$V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V$

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