Resistance is due to a "drag force" on the electrons moving through a material. This drag force arises from the electrons bumping into atoms as they move through the material. This force is due to the atoms in the material, not to the physical dimensions of the material (length and cross-sectional area).
The resistance of an object is due to its dimensions. If we make the object twice as long, the resistance will double; if we make the cross-section twice as big, the resistance will be half. If we factor out the physical dimensions, then we are left with something called the resistivity. The resistivity, r, is a property of the material. Resistance and resistivity are related by:
Calculate the diameter of a 2.0cm length of tungsten filament in a small lightbulb if its resistance is 0.050W.
Use R = r l/A, where A = pr², and the diameter, d = 2r.
Solving for A, A = lr/R.
From Table 17.1 we see that tungsten has a resistivity, r = 5.6×10-8Wm.
Thus,
A = (0.020m)(5.6×10-8Wm)/(0.050W) = 2.24×10-8m²
The diameter is therefore:,br>
d = Sqrt{4A/p} = Sqrt{4(2.24×10-8m²)/p} = 1.7×10-4m = 0.17mm.
Earlier I said that, for many materials, resistance is roughly constant over a wide range of voltage, current, temperature, humidity, etc. While it is roughly true, for most materials it is not exactly true. Normally the resistance and resistivity vary somewhat depending on conditions, but especially with temperature.
Because the variation is small, it is approximately linear. The change of resistivity with temperature is approximately:
The relationship between resistivity and resistance allows us to write a similar approximation for resistance:
A wire 3.00m long and 0.450mm² in cross-sectional area has a resistance of 41.0W at 20.0°C. If its resistance increases to 41.4W at 29.0°C, what is the temperature coefficient of resistivity?
This problem is an example where more information is given than is needed to solve the problem. Because the length and cross-sectional area are given, you may be tempted to start by calculating the resistivity in each case, but this is not necessary because the temperature coefficient for change of resistance is the same as for resistivity. Simply use the relation for the change of resistance and solve the problem directly.
I'll use R = R0[1 + a(T - T0)], and solve for a.
a = (R/R0 - 1)/(T - T0) = ((41.4/41.0) -1)/(29.0 - 20.0°C) = 0.00108/°C = 1.08×10-3/°C.
(a) A 34.5m length of copper wire at 20°C has a radius of 0.25mm. If a potential difference of 9.0V is applied across the length of the wire, determine the current in the wire. (b) If the wire is heated to 30.0°C and the 9.0V potential difference is maintained, what is the resulting current in the wire?
(a) The description of the wire is enough information for us to determine the resistance, and then we can use Ohm's law to determine the current for V = 9.0V.
For copper, r = 1.7×10-8Wm, from Table 17.1.
Determine the resistance with
R = rl/A = (1.7×10-8Wm)(34.5m)/p(2.5×10-4m)² = 3.0W.
Use Ohm's law to find the current:
I = V/R = 9.0V / 3.0W = 3.0A.
(b) The resistance increases when the wire is heated.
From Table 17.1, the temperature coefficient for copper is a = 3.9×10-3/°C.
Let R0 = 3.0W at T0 = 20.0°C, as found in part (a), then at T = 30.0°C,
R = R0[1+a(T - T0)] = (3.0W)[1 + (3.9×10-3/°C)(30.0 - 20.0°C)] = 3.1W.
The current decreases to
I = V/R = (9.0V)/(3.1W) = 2.9A.
Note: In the second paragraph, the text should read "As the charge moves from A to B through the battery, its electrical potential energy increases by the amount DQDV and ..."
As charge moves around an electrical circuit, energy is transferred from a battery (source of voltage) to resistors. Since the current is moving continuously (it is not a one time occurrence), the rate of energy transfer is a more useful quantity. The rate of energy transfer in a circuit is called the power, P, and is given by