Question

An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance.
(a) What is the current to the motor?
(b) What voltage is applied to it?
(c) What power is supplied to the motor?
(d) Repeat these calculations for when the battery connections are corroded and add 0.0900Ω to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)

Answer 1

Answer:

Explanation:

Resistance of motor, R = 0.05 ohm

internal resistance of battery, r = 0.01 ohm

Voltage of battery, V = 12 V

(a) Total resistance, R' = R + r = 0.05 + 0.01 = 0.06 ohm

Let the current be i.

use Ohm's law

i = V / R'

i = 12 / 0.06 = 200 A

(b) Voltage across motor, V' = i x R = 200 x 0.05 = 10 V

(c) Power, P = i²R = 200 x 200 x 0.05 = 2000 Watt.

(d) Total resistance, R' = 0.05 + 0.1 + 0.09 = 0.15 ohm

i = V / R' = 12 / 0.15 = 80 A

V' = i x R = 80 x 0.05 = 4 V

P' = i²R = 80 x 80 x 0.05 = 320 Watt

Answer 2

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   $V = \epsilon - Ir$

now,

   $IR = \epsilon - Ir$

   $I(R+r) = \epsilon$

   $I= \dfrac{\epsilon}{R+r}$

inserting the values

   $I= \dfrac{12}{0.05+0.01}$

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 $I= \dfrac{\epsilon}{R+r}$

   $I= \dfrac{12}{0.14+0.01}$

     I = 80 A

     V = 80 x 0.05 = 4 V

     P = 4 x 80 = 320 W