Answered

When the lights of a car are switched on, an ammeter in series with them reads 12.0 A and a voltmeter connected across them reads 11.5 V. When the electric starting motor is turned on, the ammeter reading drops to 9.00 A and the lights dim somewhat.If the internal resistance of the battery is .0500 Ohms and that of the ammeter is negligible what are the a) emf of the battery and b) the current through the starting motor when the lights areon?

Answer :

lublana

Answer with Explanation:

We are given that

Current,I=12 A

V=11.5 V

I'=9 A

Internal resistance,r=0.05 ohm

a.We have to find the emf of battery.

[tex]E=V+Ir[/tex]

Using the formula

[tex]E=11.5+12(0.05)=12.1 V[/tex]

b.Voltage of ligh

[tex]V'=I'R_{light}=9\times \frac{11.5}{12}=8.6 V[/tex]

[tex]I_{motor}=\frac{E-V'}{r}-9=\frac{12.1-8.6}{0.05}-9=61 A[/tex]

Hence, the current through the starting motor when the light areon=61 A

Other Questions