# Why do resistors always cause a voltage drop equal to the voltage of the battery

Consider a wire connected to a battery. Now,potential is analogous to the energy of the particles.And potential in a resistor drops because of the friction inside the resistor(considering there is no friction along the wire and outside the resistor).So the friction determines how much energy is lost.So why does the voltage drop to zero when the current passes from the last resistor it encounters along the wire? I mean,if there are two resistors and you have a battery of 6V,why do the resistors ALWAYS drop the voltage from 6V to 0V?Each resistor should cause a voltage drop analogous to the loss of energy due to the friction.I mean that,sometimes I(the current) should become zero(friction in resistors cause more friction than 6V-considering the convention from voltage to work W=q*ΔV) or the voltage will not reduce to zero(friction in resistors cause energy loss less than 6V).

## 1 Answer

I think the key thing missing in your thinking is that the energy drop across a resistor is not just determined by the properties of the resistor, but also by how much current flows. The cool thing is that no matter what resistors you put in, the current that flows is such that the potential will fall all the way back down. The reason for this is that electromagnetism is a conservative force, and that means that if you go all the way around you have to get back to where you started.

Think of it like a roller coaster that goes up and down hills--if the roller coaster starts at the top of a hill, rolls down without friction, and comes back around, it should get back to the top of the hill without any "extra" speed. Otherwise the rollercoaster could just go around again and again and get more speed for free with no input! In this analogy, resistors don't work like a friction force, they work more like the hills. You can put a charge on top of a hill (a high voltage place) and it runs out of steam (gets to 0 V) right before it is pushed back "up the hill" (i.e. crosses the voltage source)

• You analyze this macroscopically.I am talking about a microscopic analysis where resistors are considered to be "places" where charges crash on particles inside the resistor and the "walls" of the resistors.So the crashes would cause a drop in energy.And i am asking WHY that drop in energy is equal to the potential of the battery? – TheQuantumMan Mar 17 '15 at 23:51
• Think of it dynamically. The energy dissipated in a resistor will be proportional to the current already in the resistor. If the energy loss is less than the voltage, extra energy will build up and there will be more current until that cancels out. If the energy loss is more than the voltage, the current will decrease. $\delta E = V$ is the equilibrium. – zeldredge Mar 18 '15 at 0:02
• so δE=V is the steady-state.But pre-steady state,i can not understand very well what you are saying :/ .The first time that the current passes through the wire(and resistors),a drop in energy due to the resistors will take place(and it is proportional to the crashes of the charges inside the resistors).So why is that energy drop the same as the Voltage difference of the battery the FIRTS LOOP? – TheQuantumMan Mar 18 '15 at 0:09