created at May 25, 2021

Ohm's law for a full circuit

Consider a basic electric circuit with a real voltage source and load, which is represented with resistor R. The real voltage source has also its own inner resistance R0.

The current I is:

I=ER+R0I = \frac{E}{R + R_{0}}

The voltage U is not the same as electromotive force E, because some portion of voltage drops at R0:

U=EU0U = E - U_{0}

Because U is also I*R, we can compute it also this way:

U=ERR+R0U = \frac{E\cdot R}{R + R_{0}}

The power consumed by load is:

P=UIP = U\cdot I

or, expressed through the consumer's resistance R:

P=(ER+R0)2RP = \left ( \frac{E}{R + R_{0}} \right )^{2}\cdot R

Consider E = 12 V, and R0 = 1.5 Ω. Let's graph the power dependence on consumer resistance R:

When resistance is 0, we have a short circuit. All power is dissipated on the source R0. When resistance tends to infinity, we have an idle circuit, and the power tends to zero, since the total resistance is very high. The maximum possible power is dissipated on the load when load resistance R is equal to source resistance R0. However, that doesn't mean that such a scheme is the most efficient. Efficiency will be 0.5 in that case, since the half of the total power is dissipated on the load. Let's define efficiency as:

η=RR+R0\eta = \frac{R}{R + R_{0}}

And let's graph it as a function of load resistance R:

The best way to increase efficiency is to reduce the inner resistance of the voltage source.