# 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).

• 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