SAT Physics Subject Test
Chapter 10 Direct Current Circuits
Let’s say we had a copper wire and a glass fiber that had the same length and cross-sectional area, and that we hooked up the ends of the metal wire to a source of potential difference and measured the resulting current. If we were to do the same thing with the glass fiber, the current would probably be too small to measure, but why? Well, the glass provided more resistance to the flow of charge. If the potential difference is ∆V and the current is I, then the resistance is
Notice that if the current is large, the resistance is low, and if the current is small, then resistance is high. The ∆ in the equation above is often omitted, but you should always assume that in this context, V = ∆V = difference in electric potential, also called voltage.
The resistance of an object depends on two things: the material it’s made of and its shape. For example, again think of the copper wire and glass fiber of the same length and area. They have the same shape, but their resistances are different because they’re made of different materials. Glass has a much greater intrinsic resistance than copper does; it has a greater resistivity. For a wire of length L and cross-sectional area A made of a material with resistivity ρ, resistance is given by
The resistivity of copper is around 10–8 Ω • m, while the resistivity of glass is much greater, around 1012 Ω • m.
Because resistance is voltage divided by current, it is expressed in volts per amp. One volt per amp is one ohm (Ω, omega). So, 1 V/A = 1 Ω.
1. If a voltage of 9 V is applied between the ends of the wire whose resistance is 0.2 Ω, what will be the resulting current?
Here’s How to Crack It
From I = V/R, we get