# In a circuit, if voltage is established at both terminals, then how can components use up this voltage? [duplicate]

Suppose we use a 9V battery. Then the voltage between these two terminals is 9V. In other words, it would take 9 J of energy to move +1C of charge from the (+) to the (-) terminal by the electric field established by the battery.

So exactly how can electrical components "reduce" this voltage? How can we say that the potential difference between a resistor connected to a battery is 9V? This would obviously imply that the voltage would have to "move" in the circuit, such that right before the resistor, the potential is 9 V. Then after the current passes the resistor, the potential at this point immediately after the resistor, it is somehow 0 V. But this is wrong by definition, because it is +9V at the + terminal of the battery, and 0V at the - terminal of the battery.

I am confused?

## marked as duplicate by Brian Moths, Kyle Kanos, Kyle Oman, Qmechanic♦Feb 21 '15 at 1:24

1) When a charge $q$ is moving (say at velocity v) through a perfect conductor such as an ideal wire, it requires no force to maintain its velocity because it encounters no resistance. This is good, since there can be no electric field inside a perfect conduct and thus no force can be applied to the charge by the electric field.
3) Because resistors are not perfect conductors there is a "resistive" force (say, $-\alpha v$) on the charge and thus it requires a force on the charge to keep it moving along. This force must be applied by an electric field which has built up within the resistor, which requires a drop in voltage when passing through the resistor of IR (it is linear in the current because the resistive force is linear in the velocity).
So, in your example, the fact that the battery is 9 volts means that the battery is saying "I will do $9q$ work on each charge to get it through the loop". But, for example, if there is only one resistor R, the battery only needs to do $\alpha v L$ work on the charge, where L is the length of the resistive segment. Thus, the velocity of the charge in the loop will adjust to make the work come out right. I.e., the current (which is $qnv$) will adjust to become I=V/R.