Ask Question
14 January, 06:37

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.)

+3
Answers (1)
  1. 14 January, 06:42
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers