Know that electrical energy is transferred to heat energy and other forms of energy in the resistor, or other circuit components, and then into the surroundings

4.2.4 Resistance – Energy Conversion

Objective

To understand how electrical energy is converted into heat (and, where relevant, light or other forms) in a resistor or any circuit component, and how that energy is transferred to the surroundings.

1. What is resistance?

  • Resistance (R) is the property of a material that opposes the flow of electric charge. Its unit is the ohm (Ω).

  • For a uniform conductor

    \$R = \rho\frac{L}{A}\$

    where ρ = resistivity (Ω·m), L = length (m) and A = cross‑sectional area (m²).

  • Resistance can also be obtained directly from measured voltage and current:

    \$R = \frac{V}{I}\$

2. Ohm’s law

The linear relationship between the voltage across a resistor, the current through it and its resistance is

\$V = IR\$

It holds for ohmic conductors (those whose I‑V graph is a straight line through the origin).

3. Qualitative relationships

  • Length: \(R \propto L\) – longer conductors have larger resistance.

  • Cross‑sectional area: \(R \propto \dfrac{1}{A}\) – a thicker wire has lower resistance.

  • Material (resistivity): \(R \propto \rho\). Metals have low ρ, insulators have high ρ.

  • Temperature: for most metals

    \$R = R{0}\bigl[1+\alpha\,(T-T{0})\bigr]\$

    where \(\alpha\) is the temperature‑coefficient of resistance.

4. Practical determination of resistance

Standard set‑up (voltmeter‑ammeter method)

  1. Connect the resistor in series with an ammeter and a variable DC source.

  2. Place a voltmeter in parallel across the resistor.

  3. Adjust the source, record several pairs of \((V, I)\) values.

  4. Plot \(V\) (vertical) against \(I\) (horizontal).

    • The graph should be a straight line for an ohmic resistor.

    • The slope of the line equals the resistance \(R\) (since \(V = IR\)).

  5. Alternatively, calculate \(R\) for each pair using \(R = V/I\) and take the average.

5. Current‑voltage (I‑V) characteristics

ComponentTypical I‑V shapeWhat the slope tells you
Ideal resistor (ohmic)

V

I

Straight line through the origin; slope = \(1/R\).
Filament lamp

V

I

Curve is concave upwards; resistance increases as the filament heats.
Diode (non‑ohmic)

V

I

Very little current until a “knee” voltage is reached; slope varies, so a single R cannot describe it.

6. Energy conversion in a resistor

  • When a current \(I\) flows through a resistance \(R\), electrical energy supplied by the source is dissipated inside the component.

  • The dominant form of this dissipated energy is heat (Joule heating). In some devices (e.g., filament lamps) part of the energy is emitted as visible light and infrared radiation.

7. Power dissipated in a resistor

The rate at which electrical energy is turned into heat is the power \(P\) (watts, W). Using Ohm’s law, three equivalent expressions are available:

FormulaWhen it is most convenient
\(P = VI\)Both voltage across and current through the component are known.
\(P = I^{2}R\)Current and resistance are known.
\(P = \dfrac{V^{2}}{R}\)Voltage and resistance are known.

8. Energy transferred to the surroundings

If the resistor operates for a time \(t\) (seconds), the total energy released as heat is

\$E = Pt\$

With \(P\) in watts and \(t\) in seconds, \(E\) is obtained in joules (J).

9. Factors that affect the amount of heat produced

  • Current (\(I\)): \(P \propto I^{2}\). Doubling the current raises the heat fourfold.

  • Resistance (\(R\)):

    • For a given current, \(P \propto R\).

    • For a given voltage, \(P \propto 1/R\).

  • Time (\(t\)): Longer operation transfers proportionally more total energy.

  • Material & geometry: Resistivity, length and cross‑section determine \(R\) (see Section 1).

  • Temperature: An increase in temperature usually raises \(R\) (via the temperature‑coefficient), which in turn changes the heat produced.

10. Example calculations

  1. Power and energy in a resistor

    A 10 Ω resistor carries a current of 2 A for 5 minutes.

    • Power: \(P = I^{2}R = (2\ \text{A})^{2}\times10\ \Omega = 40\ \text{W}\)

    • Time: \(t = 5\ \text{min}=300\ \text{s}\)

    • Energy: \(E = Pt = 40\ \text{W}\times300\ \text{s}=1.2\times10^{4}\ \text{J}\)

  2. Resistance of a copper wire

    Length \(L = 2\ \text{m}\), cross‑sectional area \(A = 1.0\times10^{-6}\ \text{m}^{2}\), resistivity \(\rho = 1.68\times10^{-8}\ \Omega\!\cdot\!m\).

    • Resistance: \(R = \rho L/A = \dfrac{1.68\times10^{-8}\times2}{1.0\times10^{-6}} = 0.0336\ \Omega\)

    • If a current of 3 A flows, power dissipated:

      \(P = I^{2}R = (3\ \text{A})^{2}\times0.0336\ \Omega = 0.302\ \text{W}\)

  3. Why a filament lamp gets hotter than a metal heating element of the same resistance

    The filament is made of a material (e.g., tungsten) with a very high temperature‑coefficient of resistance and low thermal conductivity. When the same voltage is applied, its resistance rises sharply as it heats, allowing a much higher temperature to be reached. Consequently a larger fraction of the electrical energy leaves as visible light and infrared radiation, whereas a typical metal heating element stays cooler because its resistance changes only slightly with temperature.

  4. Power in a non‑ohmic device (e.g., a diode)

    Read the voltage \(V\) and current \(I\) at the required operating point from the I‑V graph, then use \(P = VI\). Because the relationship is not linear, you cannot use \(I^{2}R\) or \(V^{2}/R\) unless you first determine an “effective” resistance at that point (the local slope of the curve).

  5. Resistance of a nichrome wire at a higher temperature

    Given \(R{0}=15\ \Omega\) at \(T{0}=20^{\circ}\text{C}\) and \(\alpha =4.0\times10^{-3}\ \text{°C}^{-1}\), find \(R\) at \(T=120^{\circ}\text{C}\).

    • \(R = R{0}\bigl[1+\alpha(T-T{0})\bigr] = 15\bigl[1+4.0\times10^{-3}(120-20)\bigr]\)

    • \(R = 15\bigl[1+4.0\times10^{-3}\times100\bigr]=15(1+0.40)=21\ \Omega\)

11. Summary of key points

  • Resistance opposes charge flow; \(R = \rho L/A\) and \(R = V/I\). It increases with length and temperature, and decreases with cross‑sectional area.

  • Ohm’s law (\(V = IR\)) describes ohmic conductors; the slope of an I‑V graph gives the resistance.

  • Electrical energy is mainly converted into heat in a resistor; in devices such as filament lamps part of the energy appears as light.

  • Power dissipated can be calculated with any of the three equivalent formulas: \(P = VI\), \(P = I^{2}R\) or \(P = V^{2}/R\).

  • Total energy transferred to the surroundings is \(E = Pt\).

  • Heat production grows with the square of the current and depends on resistance, time, material, geometry and temperature.

12. Practice questions

  1. A 5 Ω heater is connected to a 12 V supply. Calculate:

    • The current through the heater.

    • The power it dissipates.

    • The energy released after 10 minutes.

  2. A copper wire (resistivity \(1.68\times10^{-8}\ \Omega\!\cdot\!m\)) is 2 m long and has a cross‑sectional area of \(1.0\times10^{-6}\ \text{m}^{2}\). Find its resistance and the power dissipated when a current of 3 A flows through it.

  3. Explain why a filament lamp gets hotter than a metal heating element of the same resistance when both are connected to the same voltage source.

  4. For a non‑ohmic device (e.g., a diode) the V‑I curve is not a straight line. Briefly describe how you would determine the power dissipated at a particular operating point.

  5. A nichrome wire has a resistance of 15 Ω at 20 °C and a temperature‑coefficient \(\alpha = 4.0\times10^{-3}\ \text{°C}^{-1}\). What is its resistance at 120 °C?