Answered

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0790 Ω. The headlights together have an equivalent resistance of 4.00 Ω (assumed constant). (a) What is the potential difference across the headlight bulbs when they are the only load on the battery? (Enter your answer to at least two decimal places.) V (b) What is the potential difference across the headlight bulbs when the starter motor is operated, with 35.0 A of current in the motor? (Enter your answer to at least two decimal places.)

Answer :

Answer:(a) 0.24v (b) 140.00v

Explanation:I= EMF/R+r= 12.6v/4.00+0.0790 =3.0890A

v=IR=3.0890*4.00=12.35v

V=EMF-v=12.6v-12.356=0.24v

hence potential difference across the headlight bulbs when they are the only load (V) =0.24volts.

(b) v=IR=35.0*4.00=140.00 volts

Hence the potential difference across the headlight bulbs when the starter motor is operated (V) = 140.00volts

Other Questions