An automobile battery has an emf of 12.6 v and an internal resistance of 0.0570 ω. the headlights together have an equivalent resistance of 3.30 ω (assumed constant). (a) what is the potential difference across the headlight bulbs when they are the only load on the battery

ΔV =12.38 V

Explanation:

Given that

V= 12.6 V

r= 0.057 Ω

R= 3.3Ω

The equivalent resistance

R'=r+R

R'=3.3 + 0.057

R'=3.357 Ω

We know that

V= I R'

I =Current

12.6= I x 3.357

I=3.375 A

The voltage difference across the headlight bulbs

ΔV = I R

ΔV  = 3.375 x 3.3

ΔV =12.38 V

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