An automobile battery has an emf of 12.60 V and an internal resistance of 0.078

Question

An automobile battery has an emf of 12.60 V and an internal resistance of 0.078 0 ?. The headlights together have an equivalent resistance of 3.800 ? (assumed constant). (a) What is the potential difference across the headlight bulbs when they are the only load on the battery? (Give your answer to four significant figures.)
1 V

(b) What is the potential difference across the headlight bulbs when the starter motor is operated, requiring an additional 35.0 A from the battery? (Give your answer to four significant figures.)
2 V (a) What is the potential difference across the headlight bulbs when they are the only load on the battery? (Give your answer to four significant figures.)
1 V

(b) What is the potential difference across the headlight bulbs when the starter motor is operated, requiring an additional 35.0 A from the battery? (Give your answer to four significant figures.)
2 V

Explanation / Answer

(a) Total resistance in circuit = 3.80 + 0.078 = 3.878 ohms Voltage across 3.8 ohms = 12.6 V *(3.80) / (3.878) Voltage across headlights = 12.38 V (b) The additional voltage drop across the battery internal resistance is: V = (35A)(0.078 ohms) = 2.73 V Therefore, the headlight voltage is now = 12.38 V - 2.73 V = 9.65 V

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