When the lights of an automobile are switched on, an ammeter in series with them reads 12 A and a voltmeter connected across them reads 15 V. See Fig. 28-52. When the electric starting motor is turned on, the ammeter reading drops to 8.0 A and the lights dim somewhat. If the internal resistance of the battery is .0500 Ohms and that ofthe ammeter is negligible what are the

a) emf of the battery and

b) the current through the starting motor when the lights areon? Thank you so much for your help.

A) E.M.F = 15.6V

B) Motor current = 104A

Explanation:

A) From the question, we see that when lights are on (but not motor), the circuit current = 12 A

Now, we know that V= IR where V = 15V in the question.

Thus, R = V/I

The resistance of the lights is given by; R = 15/12 = 1.25 ohms

The voltage drop across the internal resistance = Ir

From the question, internal resistance = 0.05 ohms.

Thus,

The voltage drop across the internal resistance = 12 x 0.05 = 0.6V

Hence,

Battery emf = 15 + 0.6 = 15.6V

B) From the question, there's an assumption that the resistance of the lights does not change when the lights dim.

Thus, when starter motor is on,

Current through 1.25 Ω lights is now 8A.

Thus, the voltmeter reading is now;

v = ir = (8A * 1.25 Ω) = 10 V

voltage across internal resistance of battery = ( emf - 10V) = 15.6V - 10V = 5.6V

Also, current though internal resistance of battery is;

i = v/r = ( 5.6V / 0.05 Ω ) = 112 A

Now, from the question, the 8A current is going through the lights and the rest is going through the starter motor.

Thus;

motor current = (112A - 8 A) = 104A

answer: a. same unit

explanation:

amplitude

hope this