This equation tells us that we can maximize the current by having a large voltage drop
and a small resistance. This is one of the most important equations dealing with
electromagnetism, and SAT II Physics is bound to call upon you to remember it.
EXAMPLE
Three batteries are added to a circuit, multiplying the potential difference in the circuit by
four. A resistor is also added, doubling the resistance of the circuit. How is the current in
the wire affected?Taking the initial voltage to be V and the initial resistance to be R, the initial current is
= V/R. The new voltage is 4 V and the new resistance is 2 R, so the final current is:
These changes double the current.
Resistivity
Resistivity, , is a property of a material that affects its resistance. The higher the
resistivity, the higher the resistance. Resistance also depends on the dimensions of the
wire—on its length, L, and cross-sectional area, A:
A longer wire provides more resistance because the charges have farther to go. A larger
cross-sectional area reduces the resistance because it is easier for the charges to move.
The unit of resistivity is the ohm-meter, · m. The resistivity of copper is about 10 –8 ·
m and the resistivity of glass is about 1012 · m. At higher temperatures, the resistivity of
most metals increases.
EXAMPLE
A copper wire of length 4 m and cross-sectional area 4 mm^2 is connected to a battery with a
potential difference of 9 V. What is the current that runs through the wire? Approximate the
resistivity for copper to be 10–8 · m.As we know, the current in a wire is a measure of voltage divided by resistance. We know
that the voltage for the circuit is 9 V, but we don’t know the resistance. However, since we
know that the resistivity for copper is 10 –8 · m, we can use the formula for resistivity to
calculate the resistance in the wire.
First, we need to remember that area is measured in m^2 , not mm^2. If 1 mm = m,
then 4 mm^2 = = m^2.