# In a DC circuit with one battery and one resistor, why does the voltage drop remain the same irrespective of the value of the resistor?

DC Circuit#1: I have a 10V battery connected to a 10 Ohm Resistor. Therefore, the current will be 10/10 = 1 Amps

DC Circuit#2: I have the same 10V battery connected to a 5 Ohm Resistor. Therefore, the current will be 10/5 = 2 Amps

(Assume the conductor wire does not have any resistance at all)

What I do know: The negative terminal of the battery produces a repulsive force on the electrons. This force travels at near the speed of light. When the electrons enter the resistor, they start colliding with the lattice. The kinetic energy of the electrons are converted to heat energy which is dissipated. The drift velocity of the electrons slow down inside the resistor. Now this slowing down effect is propagated backwards because of which the entire drift velocity through the whole of the circuit attains a steady state. In Circuit#1, this steady state drift causes a charge flow per unit time, of 1A. In Circuit#2, the resistance is less, so the steady state drift is more, so the charge flow per unit time is 2A. Because the whole of the circuit has the same steady state, an ammeter connected anywhere in the circuit shows the same Ampere reading. I completely get this 'same current everywhere thing'. My question is more about the potential drop.

My question: In Circuit#1, the resistance is more (10Ohm). This means that the electrons encounter more collisions while travelling through the resistor. So the electrons lose much energy by the time they come out of the resistor. Compare this to Circuit#2, the resistance is less. This means that the electrons collide less. Agreed that the electrons lose energy while travelling through the resistor, in Circuit#2 also. But my point is that if you compare the electrons at the point of exit point of the resistors, between Circuit#1 and Circuit#2, the electrons in Circuit#1 have lost much more of their ability to do work because they were slowed down more by the collisions, than the electrons in Circuit#2. Then why ,in the whole world, is the potential difference measured across the resistor the same(10V), in both the circuits?

If you keep on increasing the resistance, I understand that the current will keep on decreasing. But why doesn't the potential difference between both ends of the resistor change? The energy difference of the electrons at the entry point of the resistor, and the electrons existing the resistor will definitely depend on what was the amount of collisions that happened in the journey through the resistor. So why doesn't the potential difference also depend on this?

• Voltage is the Potential to supply Energy but real energy is VI*t and depends on V^2/R for power – Tony Stewart Sunnyskyguy EE75 Apr 21 '18 at 17:57
• Take a look at this : en.wikipedia.org/wiki/Voltage_divider – Long Pham Apr 21 '18 at 17:57
• Because in this ideal circuit, the voltage is defined by the voltage source, and the voltage source alone. – Dampmaskin Apr 21 '18 at 18:05
• "why doesn't the value of the resistance affect the potential difference." - this because you consider the ideal voltage source, which is, by definition, always supply 10V no matter which load is there. You seem to be confused with the concept of ideal voltage source. Then consider a more realistic model, which includes internal resistance of a battery. – Ale..chenski Apr 21 '18 at 18:05
• Weird that the question seems like - "I apply a 50N force on a block of mass. So why the force felt by the block is 50N ?" – Meenie Leis Apr 22 '18 at 16:07

"why doesn't the value of the resistance affect the potential difference."

• this is because you consider the ideal voltage source, which, by definition, always supplies 10V (in your example) no matter which load is there, or how "hard" electrons do their work scattering over phonons in the resistor lattice. To account for "hard work" of electrons and a drop of potential, consider a more realistic model, which includes internal resistance of a battery and essentially models the hard work of electrons in realistic power source/battery.
• Say I have a 10V battery. Just assume that the battery is a new one, and that the resistance we are talking about is a miniscule one. Still my point is why the potential drop across is 10V across the resistor. A miniscule resistor means that the electrons only have done very very less work by the time they exit the resistor. So the energy level difference of electrons between entry and exit points of the miniscule resistance is almost nil. But still I get a P.D. of 10V across it. This is what baffles me! – Sujith George Apr 21 '18 at 18:30
• If the voltage drop was not 10V, the voltage source would not have been 10V. If you're honestly baffled by the fact that 10V is equal to 10V, it might be time to double check your assumptions. You yourself have explicitly defined the voltage to be 10V in both your examples. Then you proceed to ask why the voltage in both examples is 10 V. The simple answer is that it is 10V because you yourself defined it to be 10V. Regardless of energy, electrons, resistance and anything else. – Dampmaskin Apr 21 '18 at 19:13
• @Dampmaskin, this was a very good assessment of yours. Thanks. – Ale..chenski Apr 21 '18 at 19:26
• @SujithGeorge, minscule or not, this is the internal resistance that will define your change in voltage. Every battery has the limit how much current it can generate, which has a model of "internal resistance". All your considerations about electron scattering are in fact the foundations of resistance concept - more scattering - more resistance - less current at fixed applied potential difference. Try to do some simple experiments with an AA alkaline battery and various resistors, and use DMM to take voltage measurements. You will see the effect of internal resistance immediately. – Ale..chenski Apr 21 '18 at 19:33
• @Dampmaskin I see what you are getting at. So I have said that my battery always pumps out a voltage of 10V no matter what. Agreed. So I understand why the potential at the 'entry' point of the resistor is always high, thanks to my ideal battery. But the confusing part is, when the electrons exit the resistor and reach the +ve terminal of the battery, they still have 'some' potential left in them. Agreed, they have made some collisions, but they are not exhausted completely, by the time they reach the battery terminal. So why is the 'difference' between the entry and exit point still 10V? – Sujith George Apr 21 '18 at 20:41

Potential Difference between two points a and b is defined as the work done per charge to move from a to b.

$$\Delta P = W/Q$$

Assuming that voltage source is ideal and has no internal resistance. A resistor is connected between the voltage source.

The net amount of charges flowing in both circuits and the net work done varies in both circuits. But let's consider only the movement of a unit charge for the sake of simplicity in explaining potential difference.

$$\implies \Delta P = W$$

So it means Potential Difference is the work done by a unit charge to move from one terminal to the terminal of voltage source.

If it is a 10V battery, it means 10 joules have to be spent by/on a 1C charge to travel from one terminal to the other terminal through a conductor. Work is a state function. It means, it is not dependent on the path taken. It depends only on the end points of the energy source, which is the battery here. So in both Circuit#1 and Circuit#2 , whether one case has more resistance or collisions in path or whatever, in the end an electron will have to do the same amount of work or should have lost the same amount of energy to reach the -ve terminal from the +ve terminal.

It is analogous to a person climbing a slope and stairs. He will have to do the same amount of work against the gravity to reach the same height. Even though the way he reached there was different.

That's the whole point of potential difference, work per unit charge stays the same.