On a cold morning in Lucknow or Delhi, many homes drop an immersion rod into a bucket of water for the bath. The rod is a sealed metal tube with a coil inside, and nothing in it moves or burns. Within twenty minutes the water is hot. Every joule of that heat came from charges drifting through the coil.
Where the energy goes
A steady current \(I\) flows through a resistor \(R\), with \(V = IR\) across it. In a time \(t\) a charge \(q = It\) enters at the high-potential end and leaves at the low one. It loses potential energy \(qV\). The current is steady, so the charges leave no faster than they came, and none of this energy becomes their kinetic energy.
The energy goes to the lattice. Between collisions the field speeds each electron up a little; at each collision the electron hands that energy to an ion. The ions vibrate harder, which means the resistor is hotter. This is Joule heating: electrical energy turned into heat, one collision at a time.
\[ dq = I\,dt \]
Why. the charge that crosses the resistor in a short time \(dt\).
\[ dW = dq\,V = I\,V\,dt \]
Why. a charge \(dq\) falling through a potential difference \(V\) loses energy \(dq\,V\).
\[ dH = I\,(IR)\,dt = I^2R\,dt \]
Why. with a steady current all of it becomes heat, and \(V = IR\).
\[ H = I^2 R\,t \]
Read it. add the pieces over a time \(t\) at a steady current.
The heat produced in a resistor by a steady current is proportional to the square of the current. It is also proportional to the resistance and to the time: \(H = I^2Rt\).
A \(10\,\Omega\) resistor carries \(2\,\text{A}\). Find the rate at which heat is produced and the heat produced in \(5\) minutes.
The rate is \(I^2R = 2^2 \times 10 = 40\,\text{W}\), that is \(40\,\text{J}\) each second.
In \(5\,\text{min} = 300\,\text{s}\), \(H = 40 \times 300 = 12\,000\,\text{J} = 12\,\text{kJ}\).
Values used in this lesson: \(c_\text{water} = 4200\,\text{J/(kg K)}\), \(L_f = 3.36 \times 10^5\,\text{J/kg}\) for ice, the mains at \(220\,\text{V}\) and Rs 5 per unit.
A \(3\,\Omega\) wire carries \(4\,\text{A}\) for \(7\) minutes. The wire lies in ice at \(0\,°\text{C}\). Find the heat produced and the mass of ice melted, with \(L_f = 3.36 \times 10^5\,\text{J/kg}\).
\(I = 4\,\text{A}\), \(R = 3\,\Omega\), \(t = 420\,\text{s}\). All the heat goes to melt ice at its melting point.
\(H = I^2Rt\) and \(m = H/L_f\).
\(H = 16 \times 3 \times 420 = 20\,160\,\text{J}\). Then \(m = 20\,160/(3.36 \times 10^5) = 0.06\,\text{kg}\).
The wire produces \(20\,160\,\text{J}\) and melts \(60\,\text{g}\) of ice.
The current through a resistor is doubled for the same time. How does the heat produced change?
Three forms of the power
The rate of heating is the power, \(P = H/t\). Because \(V = IR\), it has three equal forms: \(P = VI = I^2R = V^2/R\). One watt is one joule per second, and also one volt times one ampere.
The three forms point different ways, so choose the one that holds the fixed quantity. When the current is fixed, \(P = I^2R\) says a larger resistance heats more. When the voltage is fixed, \(P = V^2/R\) says a larger resistance heats less. Both are true; they answer different questions.
The graph above passes \(20\,\text{W}\) at \(2\,\text{A}\), so \(R = P/I^2 = 20/4 = 5\,\Omega\). At \(3\,\text{A}\) the power is \(3^2 \times 5 = 45\,\text{W}\). One and a half times the current gives two and a quarter times the power. The graph curves upward because \(P\) goes as \(I^2\).
A heater marked \(1000\,\text{W}\), \(200\,\text{V}\) promises \(1000\,\text{W}\) when \(200\,\text{V}\) is across it. It then draws \(I = P/V = 5\,\text{A}\), and its coil has \(R = V/I = 40\,\Omega\). The rating is a promise for one voltage. What travels with the device is its resistance.
A bulb is rated \(100\,\text{W}\), \(200\,\text{V}\). Find the resistance of its filament and the current it draws at \(200\,\text{V}\). Find its power on a \(100\,\text{V}\) supply, taking the resistance as unchanged.
\(R = V^2/P = 200^2/100 = 400\,\Omega\), and \(I = P/V = 100/200 = 0.5\,\text{A}\).
At \(100\,\text{V}\), \(P = V^2/R = 100^2/400 = 25\,\text{W}\).
Half the voltage gives a quarter of the power. A \(100\,\text{W}\), \(220\,\text{V}\) bulb behaves the same way: \(484\,\Omega\), and \(25\,\text{W}\) at \(110\,\text{V}\).
A \(60\,\text{W}\), \(240\,\text{V}\) bulb is run on \(120\,\text{V}\). Taking its resistance as fixed, find its power, in watts.
Now the Metro train of Lesson 1. Suppose it draws about \(2\,\text{MW}\) while it speeds up, a rough figure for a six-coach train. At \(750\,\text{V}\) the current is \(I = P/V = 2 \times 10^6/750\) \(\approx 2670\,\text{A}\). The same power at \(25\,\text{kV}\) would need only \(80\,\text{A}\). The heat wasted in the supply rails and cables goes as \(I^2\), so in the same conductors it would be about \(1100\) times smaller.
The unit on the bill
The joule is too small for a household. The electricity board sells energy in kilowatt-hours, which the bill calls units. One unit is the energy a \(1\,\text{kW}\) device uses in one hour: \(1\,\text{kWh} = 1000\,\text{W} \times 3600\,\text{s} = 3.6 \times 10^6\,\text{J}\). The kilowatt-hour is an energy, not a power. A \(100\,\text{W}\) bulb left on for \(10\) hours uses exactly one unit.
A \(2\,\text{kW}\) room heater running \(3\) hours a day uses \(2 \times 3 = 6\) units a day. At Rs 5 per unit that costs Rs 30 a day. Over a winter month of thirty days the heater alone adds Rs 900 to the bill.
A household runs a \(1.5\,\text{kW}\) geyser for \(40\) minutes and four \(15\,\text{W}\) LED lamps for \(5\) hours each day. How many units does it use in a day?
Fans, refrigerators and air conditioners sold in India carry a star label of the Bureau of Energy Efficiency. More stars mean less power for the same service. A five-star ceiling fan with a brushless motor uses about \(30\,\text{W}\), where an old fan used about \(75\,\text{W}\). Run for \(10\) hours a day, the new fan uses about \(110\) units a year, the old one about \(275\). At Rs 5 per unit the saving is about Rs 800 a year from one fan.
A \(1.5\)-ton air conditioner draws about \(1.5\,\text{kW}\) while it runs. Estimate the units it uses in a summer month if it runs \(8\) hours a night.
Heating water, and where the rest goes
A heater of power \(P\) running for a time \(t\) supplies \(Pt\). The water, of mass \(m\) and specific heat \(c\), takes \(mc\Delta T\) to warm by \(\Delta T\). With no losses, \(Pt = mc\Delta T\). A \(500\,\text{W}\) kettle heating \(1\,\text{kg}\) of water from \(20\) to \(70\,°\text{C}\) needs \(1 \times 4200 \times 50 = 210\,\text{kJ}\). That takes \(210\,000/500 = 420\,\text{s}\), seven minutes.
Real heaters lose some energy to the vessel and the air. The efficiency is the useful share, \(mc\Delta T/Pt\). In Fig 5.1 a \(1\,\text{kW}\) rod runs for \(7\) minutes and supplies \(420\,\text{kJ}\). The water warms from \(20\) to \(90\,°\text{C}\) and takes \(1 \times 4200 \times 70 = 294\,\text{kJ}\). The efficiency is \(294/420 = 70\,\%\).
A motor draws \(5\,\text{A}\) at \(200\,\text{V}\) and delivers \(800\,\text{W}\) of mechanical power. Find the power lost as heat, the efficiency, and the resistance of the windings.
The input is electrical, \(VI\). The output is mechanical. The difference heats the windings.
Input \(= VI\); heat \(=\) input \(-\) output \(= I^2R\).
Input \(= 200 \times 5 = 1000\,\text{W}\). Heat \(= 1000 - 800 = 200\,\text{W}\). Then \(200 = 5^2 R\), so \(R = 8\,\Omega\).
The motor wastes \(200\,\text{W}\) as heat and is \(800/1000 = 80\,\%\) efficient. Its windings have \(8\,\Omega\).
A \(1\,\text{kW}\) kettle heats \(1\,\text{kg}\) of water from \(25\,°\text{C}\) to \(100\,°\text{C}\). Ignoring losses, how long does it take, in seconds? Take \(c = 4200\,\text{J/(kg K)}\).
Now the bucket of the winter morning. Say it holds \(15\) litres of water at \(15\,°\text{C}\), and a warm bath needs \(45\,°\text{C}\). The water needs \(15 \times 4200 \times 30 = 1.89 \times 10^6\,\text{J}\). A \(2\,\text{kW}\) rod supplies this in at least \(1.89 \times 10^6/2000 = 945\,\text{s}\), about \(16\) minutes. The energy is \(0.525\) unit, under Rs 3 at Rs 5 per unit, a little more with losses.
- Using \(P = I^2R\) to say that a larger resistance always heats more.That holds only at a fixed current. At a fixed voltage, \(P = V^2/R\) and a larger resistance heats less.
- Calling the kilowatt-hour a unit of power.It is an energy: a power in kW times a time in hours, \(3.6 \times 10^6\,\text{J}\).
- Using \(V/I\) of a motor as its winding resistance.A motor turns most of its input into work. Only the heat obeys \(I^2R\).
- Forgetting to convert minutes to seconds in \(H = I^2Rt\).With \(I\) in A and \(R\) in Ω, put \(t\) in seconds to get joules.