Consider a DC voltage source connected to a load resistance in the circuit diagram below.

It is easy to work out the power dissipated in the resistor:

P = V x I

I = V / R

P = V^{2} / R

P = I^{2} R

What happens to the power dissipated as the load resistance R approaches 0 ohms? The current approaches an infinite value, and the power dissipated approaches an infinite value! In the real world, it is not possible to build a DC voltage source that can supply infinite power and there are no connecting wires that can handle an infinite current. The DC voltage source itself will include some "internal resistance". This is represented by R_{1} in the circuit diagram below.

With the knowledge that the voltage generator has internal resistance R_{1}, the question can be asked once again. What happens to the power dissipated as the load resistance R_{2} approaches 0 ohms? This time we have a voltage source of V_{gen} supplying current to the internal resistance of the DC voltage source (R_{1}), so the power supplied is given by:

P = V_{gen}^{2} / R_{1}

This is an interesting result: all of the power is dissipated in R_{1}, so the DC voltage source itself will warm up, and no power is supplied anywhere else. Once again in the real world the connecting wiring will have a very small resistance and will dissipate some power but this can be assumed to be negligably small in comparison to that dissipated by R_{1}. When the load resistance R_{2}is not 0 Ohms, the total power dissipated by R_{1} and R_{2} is given by:

P = V_{gen}^{2} / (R_{1} + R_{2})

It is then also easy to calculate the power dissipated by R_{1} and R_{2} individually:

I = V / R

I = V_{gen} / (R_{1} + R_{2})

P_{R1} = (V_{gen} / (R_{1} + R_{2}))^{2} R_{1}

P_{R2} = (V_{gen} / (R_{1} + R_{2}))^{2} R_{2}

Now the power dissipated in the load P_{R2} can be compared to the power dissipated in the DC voltage source internal resistance P_{R1}:

P_{R2} / P_{R1} = R_{2} / R_{1}

This is another interesting conclusion, and can be expressed in words as follows:

To efficiently transfer power from a power supply to a load it is important to ensure that the load resistance is high in comparison to the power supply internal resistance.