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

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.