An automobile battery has an emf of 12.6 V and an internal resistance of 0.0710

Question

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0710 . The headlights together have an equivalent resistance of 5.90 (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.)
V

Explanation / Answer

a)

total R is 5.971
Total I = 12.6v/5.971 = 2.11 amps
Voltage across 5.90 ohms is
E = IR = 2.11a x 5.90 = 12.45 volts

b)

Current due to lights = 2.11A

additional current due to motor = 35.0 A

total current = 37.11

thus voltage across the lamps = 12.6 - 0.0710 x 37.11 = 9.65 V

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