Electricity · Lesson 4 of 9
Ohm’s Law
“Ohm’s law keeps voltage, current, and resistance in a surprisingly tidy relationship.”
• Describe the experimental relationship between current and potential difference. • State Ohm’s law and explain the constant-temperature condition. • Interpret a straight potential difference-current graph. • Calculate current, potential difference and resistance using Ohm’s law. • Explain the function of resistance and a rheostat.
Turning up a battery arrangement can make a test bulb glow more brightly because a larger potential difference can drive a larger current. The relationship is not arbitrary. For many metallic conductors kept at constant temperature, doubling the potential difference doubles the current. Ohm’s law expresses this regular behaviour.
At constant temperature, the potential difference across a conductor is directly proportional to the current flowing through it.
The property of a conductor that opposes the flow of electric charge.
For a particular ohmic conductor, the ratio V/I remains constant. This constant is its resistance. A resistance of 1 Ω allows a current of 1 A when 1 V acts across the conductor, so 1 Ω = 1 V/A. At fixed voltage, a larger resistance permits a smaller current: I = V/R.
Investigating the Relationship
Connect a nichrome wire, ammeter, key and cells in series, with a voltmeter across the nichrome wire. Change the number of cells to obtain several readings. For each setting, record V and I and calculate V/I. If temperature is nearly constant, the ratios are close and the V-against-I points lie on a straight line through the origin.
On a graph with V on the vertical axis and I on the horizontal axis, the slope ΔV/ΔI equals resistance. A steeper line therefore represents a larger resistance. If I is on the vertical axis instead, the slope is I/V = 1/R; the axes must always be checked before interpreting steepness.
Current heats a conductor. Heating can change its resistance, so a filament bulb may not give a straight line over a wide voltage range. Ohm’s law is not a universal rule for every component under every condition.
Variable Resistance
A variable resistor used to change resistance and therefore control current in a circuit.
A rheostat changes the effective length of resistance wire included in a circuit. Increasing the included length raises resistance and lowers current. It controls current without needing to change the source voltage.
Conductors and Insulators
A good conductor offers little resistance and allows substantial current. A resistor is a component chosen to provide a particular resistance. A poor conductor offers much more opposition, while an insulator has such high resistance that current is extremely small under ordinary conditions. These descriptions are relative and depend on material and conditions.
Activity
Purpose: compare currents through different components at the same potential difference. Use a low-voltage source, key and ammeter in series. Place a nichrome wire in the test gap, record the current, and then replace it in turn with a torch bulb and a higher-resistance bulb while keeping the source unchanged. Observation: the readings differ because the components offer different resistances; a larger resistance gives a smaller current at the same voltage. Conclusion: current depends on both applied potential difference and component resistance.
Problem-Solving Strategy
- Convert milliamperes to amperes before using volts and ohms.
- Identify the unknown and select V = IR, I = V/R or R = V/I.
- Substitute values with units and calculate.
- Interpret whether the current should rise or fall when V or R changes.
- For a graph, read the axes before computing the slope.
Problem
A 6 Ω resistor is connected across 12 V. Find the current.
- 1.Given: R = 6 Ω and V = 12 V. Required: I.
- 2.Use I = V/R because voltage and resistance are known.
- 3.I = 12/6 = 2 A.
- 4.A current of 2 A is reasonable because each 6 V would drive 1 A through 6 Ω.
Problem
A component carries 250 mA when 5 V acts across it. Find its resistance.
- 1.Convert current: 250 mA = 0.250 A.
- 2.Use R = V/I.
- 3.R = 5/0.250 = 20 Ω.
- 4.Check: IR = 0.250 × 20 = 5 V.
Problem
A V-I graph contains the points (0.20 A, 3.0 V) and (0.50 A, 7.5 V). Find resistance and predict current at 12 V.
- 1.Because V is plotted against I, resistance is the slope ΔV/ΔI.
- 2.R = (7.5 − 3.0)/(0.50 − 0.20) = 4.5/0.30 = 15 Ω.
- 3.The line is consistent with the origin because 3.0/0.20 = 15 Ω and 7.5/0.50 = 15 Ω.
- 4.At 12 V, I = V/R = 12/15 = 0.80 A.
- 5.The predicted point remains on the same straight line if temperature is unchanged.
Quiz
Which condition is essential when applying Ohm’s law to a conductor?
What does the slope of a V-against-I graph represent?
A 10 Ω resistor carries 0.5 A. What voltage acts across it?
At constant voltage, what happens when resistance doubles?
What is the main purpose of a rheostat?
Practice Problems
- Find the current through 24 Ω connected across 12 V. Solution: I = V/R = 12/24 = 0.5 A.
- A 2 A current flows through 7 Ω. Find V. Solution: V = IR = 2 × 7 = 14 V.
- A device carries 40 mA at 8 V. Find R. Solution: I = 0.040 A. R = 8/0.040 = 200 Ω.
- A resistor carries 0.3 A at 6 V. Predict current at 10 V if temperature stays constant. Solution: R = 6/0.3 = 20 Ω. New I = 10/20 = 0.5 A.
- Two V-I lines have slopes 5 V/A and 12 V/A. Which conductor has greater resistance and which carries more current at 6 V? Solution: Resistances are 5 Ω and 12 Ω. The 12 Ω conductor has greater resistance. Currents are 6/5 = 1.2 A and 6/12 = 0.5 A, so the 5 Ω conductor carries more current.
Key Takeaways
• Ohm’s law states V is proportional to I at constant temperature. • Resistance equals V/I and is measured in ohms. • At fixed voltage, current decreases as resistance increases. • The slope of a V-against-I graph is resistance. • Graph axes must be checked before interpreting slope. • A rheostat controls current by changing resistance. • Heating can change resistance and produce non-ohmic behaviour.