# Why is potential difference related to current i.e. flow of charge rather than a single charge in OHM's LAW?

Please try to understand it intuitively, what i mean to say is that in the below circuit,

If we say that we have a 10 volt battery and a resistor of 5 ohm, we would have a current of 2 ohm whereas when we replace the resistor with one of resistance 10 ohm, we would have current that is exactly halved i.e. 1 ampere.

We know that the potential difference across the resistance would also be 10V, that means when '1 COULOMB of charge' moves through that resistance, it loses 10 Joules of energy. Now, when we have halved the resistance, still whatever amount of charge flows through it, for every coulomb there is a loss of 10 Joules of energy. **

So why is it so that when the resistance is halved, we have more charges flowing through, but losing the same amount of energy per charge even when the resistance is different?

** (assuming wires have negligible resistance and there is no loss of energy elsewhere).

In essence what I mean to say is that, if we keep the voltage source same and change the resistance, the energy lost per unit charge remains same (even though the same charge moves through different amount of resistances), but we have increased/decreased flow of charges respectively.

Why is it so? Why is the rate of flow of charges altered when we change the resistance but not the energy lost per charge, even when the resistances are different and the same charge moves through them?

I really hope you understand what I am asking...

• Why do you have to use caps? Have you tried to put time in the whole picture? You do realize that the power being given to the circuit by the power source to the 5$\Omega$ resistor is 20W, while to the 10$\Omega$ it's 10W. And, no, I do not understand the confusion. Power source gives 20W in one case, 10W in the other. Everything is consistent. Jun 22 '20 at 14:47
• the same charge is going through the resistance of 5 ohm and 2 ohm and still the lost energy is same how? more resistance means more collision thus more energy lost...? Jun 22 '20 at 15:51

Compare the flow of charge from one terminal of a power supply to another with the flow of water through a pipe from a high tank to a low tank. The analogous quantity to potential difference (energy transferred per Coulomb) is height difference (energy transferred per kilogram weight) of water. [I'm sure you know the equation for change in gravitational potential energy: $$\Delta E_{grav}=mg \Delta h$$.]

Suppose we use a narrower pipe. The rate of flow of water will decrease, but the height difference will stay the same. For each kilogram of water going from the top tank to the bottom tank, the same amount of energy will be transferred (but more slowly). Similarly in the electrical case; if you exchange your 5 $$\Omega$$ resistor for a 10 $$\Omega$$ resistor the charge will flow at half the rate, but the energy transferred per Coulomb will be the same.

• What I can comprehend from your answer is that when resistance is increased the charge spends more time in resistor and therefore looses the same amount of energy at a lower rate. Is it correct? Jun 22 '20 at 15:59
• That's right – though we say that charge 'goes through' a resistor when charge goes in one end and comes out of the other; it doesn't have to be the very same electrons that come out as went in. Jun 22 '20 at 16:14
• Okay, thank you very much. :) Jun 23 '20 at 3:34

So why is it so that when the resistance is halved, we have more charges flowing through, but losing the same amount of energy per charge even when the resistance is different?

The potential difference, or voltage, between two points is the work required per unit charge to move the charge between the points. The potential difference in this case is fixed at the voltage of the ideal battery in the circuit. Therefore the same work per unit charge will be done to move the charge through the resistor no matter what the value of the resistor. However the rate at which work is done will not be the same.

Although when you have halve the resistance the same amount of work (energy is lost per unit charge) is done per unit charge, the rate at which the charge moves through the resistor (the current) will be doubled. That, in turn, means the rate at which energy is lost (power) also doubles. Proof:

$$P=I^{2}R$$

$$I=\frac{V}{R}$$

$$P=\frac{V^2}{R}$$

If we halve the resistance, then

$$I=\frac{V}{R/2}$$

$$P=(I^{2}R/2)=\frac{2V^2}{R}$$

or double the power.

Hope this helps.

• Thank You so much! It cleared some of my concepts. :) Jun 23 '20 at 3:35