An automobile battery has an emf of 12.60 V and an internal resistance of 0.076

Question

An automobile battery has an emf of 12.60 V and an internal resistance of 0.076 0 . The headlights together have an equivalent resistance of 4.800 (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.)
in 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.) in V.

Explanation / Answer

E = emf = 12.60 volts

r = internal resistance = 0.076 ohm

R = equivalent resistance of bulbs = 4.8 ohm

i = current through the equivalent resistance of bulbs = E/(R + r) = 12.60/(4.8 + 0.076) = 2.6 A

Potential difference across the bulbs = V = iR = 12.60 x 4.8/(4.8 + 0.076) = 12.40 volts

b)

V' = (i + i')R = (35 + 2.6) (4.8) = 180.5 volts

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