Now we can plug the values for the resistivity of copper and the length and cross-
sectional area of the wire into the equation for resistivity:
Once we know the resistance of the circuit, calculating the current involves a simple
application of Ohm’s Law:
Conductivity
Infrequently, you may come across talk of conductivity and conductance rather than
resistivity and resistance. As the names suggest, these are just the inverse of their
resistant counterparts. Saying a material has high conductivity is another way of saying
that material has a low resistivity. Similarly, a circuit with high conductance has low
resistance. Someone with half a sense of humor named the unit of conductance the mho (
), where 1 = 1.Energy, Power, and Heat
As a charge carrier moves around a circuit and drops an amount of potential, V, in time t,
it loses an amount of potential energy, qV. The power, or the rate at which it loses energy,
is qV/t. Since the current, I, is equal to q/t, the power can be expressed as:
The unit of power is the watt (W). As you learned in Chapter 4, one watt is equal to one
joule per second.
VIR and PIV Triangles
Ohm’s Law and the formula for power express fundamental relationships between power,
current, and voltage, and between voltage, current, and resistance. On occasion, you may
be asked to calculate any one of the three variables in these equations, given the other
two. As a result, good mnemonics to remember are the VIR and PIV triangles: