Graded Ohm's law and Kirchhoff's laws labs: measure current, design a resistor for a target current, check KVL and KCL, then a viva.
On the syllabus
Ohm's law: V = IR
Series and parallel resistance
Kirchhoff's voltage law (KVL) and current law (KCL)
Power in a resistor and power ratings
Nodal analysis of a two-source circuit
You will be able to
Measure voltage across and current through a resistor, and connect a meter the right way for each.
Choose a resistor for a target current and check its power rating.
Show that voltages round a loop and currents into a node sum to zero.
Solve a two-source circuit by nodal analysis and confirm it by measurement.
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Graded challenges
Values like 4.7k, 100n or 2.2u are fine. Readings are the number only, in the unit shown. Every check measures the simulation, not a bench: ideal parts within their models, no noise.
1. Measure a current with Ohm's law
Current is how much charge flows each second. A resistor sets it: the more ohms, the less current for the same voltage.
FormulaI = V / R
Worked exampleA 9 V battery across 4.7 kΩ: I = 9 / 4700 = 1.91 mA.
On the scope or meterA meter on mA in series with the resistor shows a steady number. Hovering the resistor shows the same current.
Your task
A 12 V supply drives one 2.2 kΩ resistor, R1.
Work out the current with Ohm's law first. Then measure it: put the multimeter on mA in series with R1, or read R1's current by hovering it.
Write the current in milliamps.
2. Choose a resistor for 10 mA
Design runs Ohm’s law backwards: you know the voltage and the current you want, so you solve for the resistor, then pick the nearest value you can buy.
FormulaR = V / I
Worked exampleFor 20 mA from 9 V: R = 9 / 0.02 = 450 Ω, and the nearest E24 value is 470 Ω, which gives 19.1 mA.
On the scope or meterThe current through R1 settles at once: a DC circuit with no capacitors has no transient.
Your task
A 5 V supply feeds R1 alone. Pick R1 so that 10 mA flows, within 5%.
Calculate it, then pick the nearest standard (E24) value, set it, and check the current in the simulation.
3. Series resistors and Kirchhoff's voltage law
In series there is one path, so one current. Each resistor drops a share of the supply in proportion to its resistance, and the drops add up to the supply: that is Kirchhoff’s voltage law.
FormulaI = V / (R1 + R2 + R3), and V2 = I × R2
Worked example10 V across 1 kΩ + 4 kΩ: I = 10 / 5000 = 2 mA, so the 4 kΩ drops 8 V and the 1 kΩ drops 2 V. 8 + 2 = 10.
On the scope or meterA voltmeter across each resistor in turn; the three readings add up to the supply.
Your task
R1 (1 kΩ), R2 (2.2 kΩ) and R3 (3.3 kΩ) are in series across 12 V.
Measure the voltage across R2 alone (red lead on its top, black on its bottom) and the current in the loop.
Check that the three drops add up to 12 V. That is KVL.
4. Parallel branches and Kirchhoff's current law
Parallel branches share one voltage and split the current. The current into a node equals the current out of it: Kirchhoff’s current law.
FormulaR∥ = R2 R3 / (R2 + R3), and I2 = V / R2
Worked example6 kΩ and 3 kΩ in parallel make 2 kΩ. With 4 V across them, 0.67 mA flows in the 6 kΩ and 1.33 mA in the 3 kΩ: 2 mA in total.
On the scope or meterThe current in R1 equals the two branch currents added together.
Your task
A 10 V supply feeds R1 (1 kΩ), which then splits into R2 (2 kΩ) and R3 (3 kΩ) in parallel.
Measure the current in each of the three resistors.
The current into the node (R1) must equal the currents out (R2 plus R3). That is KCL.
5. Two sources: nodal analysis
Nodal analysis: pick the unknown node voltage, write that the currents leaving the node add up to zero, and solve the one equation.
Formula(Va − V1)/R1 + (Va − V2)/R2 + Va/R3 = 0
Worked exampleWith 6 V and 3 V each through 1 kΩ into a node with 1 kΩ to ground: (Va − 6) + (Va − 3) + Va = 0, so Va = 3 V.
On the scope or meterOne voltmeter reading at node A, and the current in R2, which flows from V2 into A when V2 is above A.
Your task
V1 (10 V) feeds node A through R1 (1 kΩ). V2 (9 V) feeds the same node through R2 (2 kΩ). R3 (1 kΩ) goes from A to ground.
Write KCL at node A and solve for its voltage. Then measure the voltage at A and the current in R2.
6. Respect the power rating
A resistor turns electrical power into heat. Its rating is the most it can dissipate safely; good design stays under it with some margin.
FormulaP = V × I = V² / R = I² R
Worked example9 V across 330 Ω: P = 81 / 330 = 0.245 W, right at the limit of a 1/4 W part: too close.
On the scope or meterThe resistor’s power is its current squared times its resistance; a part run over its rating glows on the canvas.
Your task
R1 is a 1/4 W resistor across a 12 V supply. Choose R1 so that at least 15 mA flows but it turns no more than 250 mW into heat.
P = V² / R. Find the range of values that works, then pick a standard one inside it.
Viva questions
The questions an examiner asks at the bench. 75% passes; every answer comes with why.