Section 35.4 Power Dissipation
Electrons in motion collide with fixed nuclei, other electrons, and vibrational excitations called phonons. As a result, a material heats up when a current flows through it. In this context, we refer conductors as resistors. The heating of resistors is responsible for such useful applications of electricity as the conventional electric oven and the light bulb.
The rate of heating of a resistor can be calculated by using the principle of conservation of energy. Suppose current \(I\) flows through a cylindrical wire across which a voltage difference \(\Delta \phi=\phi_1-\phi_2\) is maintained (Figure 35.4.1). Let \(n\) be the number of conduction electrons per unit volume and \(v\) their average drift speed. Let \(A\) be the area of cross-section and \(R\) the resistance of the wire. As a result of the current in the wire, electrons in the wire will move a distance \(v \Delta t\) in time \(\Delta t\text{.}\)

What will be the change in energy of the segment of wire between points 1 and 2 in Figure 35.4.1 during this interval? Notice that \(nvA\Delta t\) electrons will enter the wire at potential \(\phi_2\) and leave at potential \(\phi_1\text{.}\) The entering electron at 2 will bring in energy equal to \(-envA\Delta t \phi_2\) since the potential energy of a charge \(q\) at a point of potential \(\phi\) is equal to \(qV\text{.}\) The outgoing electrons at 1 will take away energy equal to \(-envA\Delta t \phi_1\text{.}\) The difference will be increase in energy of the segment between 1 and 2. This energy heats up wire with heat equal to
Therefore, the rate of heating of the wire segment, also called power dissipated in the wire, will be given by dividing heat by time. This gives
The quantity \(e n v A\) is the rate of net charge flow in the wire, which is the current \(I\) in the wire. The difference \(\left( \phi_1 - \phi_2 \right)\) is the potential difference \(\Delta\phi\text{,}\) which is also denoted by voltage symbol \(V\text{.}\)
By using Ohm's law \(V = I R\) we can rewrite power in alternate forms.
The SI unit of power is Watts (\(\text{W}\)), which is same as Joule/sec (J/s). In electrical units, \(1\text{ W}\) of power is equal to \(1\text{ VA}\text{,}\) or, \(1 \text{ A}^2\Omega\text{,}\) or, \(1 \text{ V}^2/\Omega\text{.}\)
Checkpoint 35.4.2. Energy Used by Light Bulb.
A \(60\)-W light bulb has \(120\)-V drop across its terminals. (a) How much energy is used by the bulb in \(10\) hours? (b) Determine the resistance of the filament. (c) Determine the height \(H\) that you could lift a \(50\)-kg person with the same amount of energy.
Use definitions.
(a) \(2.2\times 10^6\ \text{J} \text{,}\) (b) \(4.2\times 10^{-3}\Omega \text{,}\) (c) \(4.4\times 10^3\ \text{m}\text{.}\)
(a) Energy Used = \(P\Delta t\) = \(60\ \text{W}\times 10\ \text{h}\times 3600\ \text{s/h}\) = \(2.2\times 10^6\ \text{J}\text{.}\)
(b) The resistance \(R\) of the filament, \(R = P/(\Delta \phi)^2 = 60\ \text{W}/(120\ \text{V})^2 = 4.2\times 10^{-3}\Omega\text{.}\)
(c) The given condition on energy gives \(mgH = 2.2\times 10^6\ \text{J}\text{.}\) Therefore, we obtain \(H = 2.2\times 10^6\ \text{J}/ \left( 50\ \text{kg}\times 9.81\ \text{m/s}^2 \right) = 4.4\times 10^3\ \text{m}\text{.}\)
Checkpoint 35.4.3. Resistance in Filaments of a 60-W Light Bulb.
(a) How much is the resistance in the filaments of a 60-W light bulb that operates at a voltage of 220 volts?
(b) If in 10 seconds bulb emits 200 J of energy in light, what is its efficiency?
(c) Supposing light is produced with the same efficiency when you use the bulb in a 110-V line, how much light energy will be generated in 10 seconds?
Use \(P = \dfrac{V^2}{R} \text{.}\)
(a) \(810 \Omega\text{,}\) (b) \(33.3\%\text{,}\) (c) \(50\text{ J}\text{.}\)
(a) The bulb uses energy at a rate of \(60\text{ W}\) at \(\Delta\phi = 220\text{ V}\text{.}\) Using the formula for power dissipation in terms of potential difference and resistance we find
Therefore,
(b) The energy emitted in desired form in 10 seconds = \(200 \text{J}\) while the energy used in 10 seconds \(= 60 \text{ W} \times 10 \text{ s} = 600 \text{ J}\text{.}\) Therefore, the efficiency is
(c) With \(R = 810\:\Omega\) and \(\Delta\phi=110\text{ V}\text{,}\) the power used will be
With 33.3% efficiency, the light energy in 10 seonds will be
Note that the light energy is not 1/2. Actually, it is 1/4. You could make the arguments based on \(P \sim (\Delta\phi)^2\text{.}\) Can you see how?
Checkpoint 35.4.4. Resistance of Heating Element from Power Consumed.
A six-inch coil electric heater uses 1500 Watts power at the high setting. If the voltage across the element is 120 V, what is the resistance of the element.
Checkpoint 35.4.5. Rise in Temperature of a Copper Rod Carrying Current.
A copper rod of length \(50 \text{ cm}\) and diameter \(4 \text{ mm}\) carries a current of \(2 \text{ A}\text{.}\) If the specific heat of copper is \(380 \text{ J/kg.K}\text{,}\) what is the rate at which its temperature rises? Use density of copper \(8.23 \text{ g/cc}\text{.}\)
Energy used in raising temperature by \(\Delta T \) is \(m c \Delta T\text{,}\) where \(m\) is the mass and \(c\) the specific heat.
\(1.36\times 10^{-4} \ \text{K/s}\text{.}\)
Let us find the rate of dissipation of energy in copper by using the power formula, \(P = I R^2\text{.}\) We need \(R\text{.}\)
Hence, power dissipated in the copper rod is
The heating of copper takes place by the following equation.
From the given data we know that mass of the rod is
Hence, the rate of rise of temperature is
where K is degrees Kelvin. Change of temperature in Kelvin equals the change in temperature in Celsius. So, the rate will be \(1.36\times 10^{-4}\ ^{\circ}\text{C/s}\text{.}\)
Checkpoint 35.4.6. Heating an Aluminum Rod by Passing a Current Through it.
An aluminum rod of length \(25 \text{ cm}\) and diameter \(2 \text{ m}\) is insulated. When a current is passed through the rod, its temperature rises at the rate of \(20 \) degrees Celsius per minute. Find the current.
Data: Density of aluminum \(2.7 \text{ g/cc}\text{,}\) specific heat aluminum \(900 \text{ J/kg.K}\text{.}\)
Use conservation of energy, \(m\:c_p\:\dfrac{\Delta T}{\Delta t} = P = I^2 R\text{.}\)
\(17.2 \text{ A}\text{.}\)
This problem is similar to the one solved above. The governing equation being the balancing of energy per unit time.
This gives
Putting the expression for R in terms of resistivity, and mass by the product of volume \((AL)\) and density \((D)\text{.}\)
The numerical values given in the problem are
Putting these numerical values in (1) gives the current, \(I = 17.2\) Amps.