Question

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.

Answer 1

Answer:

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 2

Answer:

a) Emf of battery = 15.6 V

b) Current through the starting motor = 104 A

Explanation:

The image for the question seem not to be attached but, when the starting motor wasn't connected, the lights are in series with the battery connection and when the starting motor is connected, it is in parallel with the lights.

a) For this part, we use the information when the starting motor isn't connected.

Let the emf of the battery be E = ?

The voltage across the terminals be V = 15 V

Let the internal resistance of the battery be R = 0.05 ohms

And the current in the circuit be I = 12 A

Since the internal resistance is modelled to be in series with the battery, the loop equation for this circuit will be

E = V + IR

E = 15 + (12)(0.05) = 15 + 0.6 = 15.6 V

From this information now, we can obtain the resistance of the lights.

From Ohm's law,

Voltage across the lights = 15 V

Current through the lights = 12 A

Resistance of the lights = R₀

V = IR₀

R₀ = (15/12) = 1.25 ohms

b) Now, the starting motor is connected and the resistance in the circuit is if the form

Internal resistance in series with the parallel combination of resistance of starting motor and the resistance of the lights.

Current through the lights = 8.0 A

Voltage across the lights = IR₀ = 8×1.25 = 10 V.

Since the starting motor is in parallel with the lights, they will have the same voltage across them, that is, 10 V

But the voltage across the internal resistance = (Emf of battery) - (Voltage across the parallel combination of lights and starting motor) = 15.6 - 10 = 5.6 V

Current through the internal resistance = (5.6/0.05) = 112 A

Current through the starting motor = 112 - 8 = 104 A

Hope this Helps!!!