- #1
blakmamba619
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1.An automobile battery has an emf of 12.6 V and an internal resistance of 0.0620 . The headlights together present equivalent resistance 6.00 (assumed constant). What is the potential difference across the headlight bulbs when they are the only load on the battery?
What is the potential difference across the headlight bulbs when the starter motor is operated, taking an additional 35.0 A from the battery?
2. V = IR
3. For the first part i just added up the resistance R+r = Req then dived the emf by the Req to get I = 2.07 A. Then i found the potential difference across the 56ohm resistor to be IR = 2.07A*6ohms = 12.42V. Which i am sure is correct. The second part is hard for me to understand the concept, what does it mean when it says taking an additional 35.0 A from this battery? ( I tried R = V/ R = 12.6/35A = 0.36 ohms then finding a new current but it doesn't work)
What is the potential difference across the headlight bulbs when the starter motor is operated, taking an additional 35.0 A from the battery?
2. V = IR
3. For the first part i just added up the resistance R+r = Req then dived the emf by the Req to get I = 2.07 A. Then i found the potential difference across the 56ohm resistor to be IR = 2.07A*6ohms = 12.42V. Which i am sure is correct. The second part is hard for me to understand the concept, what does it mean when it says taking an additional 35.0 A from this battery? ( I tried R = V/ R = 12.6/35A = 0.36 ohms then finding a new current but it doesn't work)