An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent resistance 5.20 (assumed constant).
(a) What is the potential difference across the headlight bulbs when they are the only load on the battery?
1 V
(b) What is the potential difference across the headlight bulbs when the starter motor is operated, taking an additional 35.0 A from the battery?

(a)

(b)

Explanation:

Electric Circuits

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

(a) The electromagnetic force of the battery is and its internal resistance is . Knowing the equivalent resistance of the headlights is , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

Solving for i

i=2.28\ A

The potential difference across the headlight  bulbs is

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

3. dark matter

empty space could also be a simple vacuum