An automobile battery has an emf of 12.60 V and an internal resistance of 0.086

Question

An automobile battery has an emf of 12.60 V and an internal resistance of 0.086 0 . The headlights together have an equivalent resistance of 4.200 (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.)

(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.)

Explanation / Answer

A)The total equivalent resistance would be:

Req = 0.086 + 4.2 = 4.286 Ohm

We know that, V = IR => I = V/R

I = 12.6/4.286 = 2.94 A

Drop will be:

V' = 2.94 x 4.2 = 12.35 Volts

Hence, V' = 12.35 Volts

B)The additional drop across internal resistor would be:

V = 35 x 0.086 = 3.01 V

So the new voltage across headlight will now be,

V'' = 12.35 - 3.01 = 9.34

Hence, V'' = 9.34 Volts

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