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Electricity & Magnetism · Ohm's Law

Voltage, Resistance and Current

Verify Ohm's Law for individual resistors, then construct series and parallel combinations to test the equivalent-resistance rules and Kirchhoff's loop and junction laws.

Theory — Ohm's Law and Resistor Networks

Charge, Current, and Voltage

Electric current (I) is the rate of flow of electric charge past a point in a circuit, measured in amperes (A). Voltage (V), also called potential difference, is the energy delivered per unit charge between two points, measured in volts (V). A battery maintains a potential difference that drives current through a closed circuit.

Units Current:   1 A = 1 coulomb / second
Voltage:   1 V = 1 joule / coulomb
Resistance:   1 Ω = 1 V / 1 A

Ohm's Law

For an ohmic resistor, the current through the resistor is directly proportional to the voltage across it. The constant of proportionality is the resistance R:

Ohm's Law V = I · R

Equivalent forms:
I = V / R     R = V / I
A graph of V (vertical) vs I (horizontal) is a straight line through the origin with slope = R.

Section I — Ohm's Law

Connect a single resistor to a variable battery. Vary the voltage and measure the current at each setting. Plot V vs I — the slope of the best-fit line is the resistance. Repeat for several resistors to verify V = IR.

Section II — Series and Parallel Networks

Combine two resistors in series, then in parallel. Measure the total current and the voltage across each resistor. Compare the measured values to the predicted equivalent resistance and verify Kirchhoff's voltage and current laws.

Equivalent Resistance — Series

Resistors in series carry the same current. The total voltage across the combination is the sum of the individual voltage drops, so the equivalent resistance adds:

Series Combination I_total = I₁ = I₂ = …    (same current through all)
V_total = V₁ + V₂ + …
R_series = R₁ + R₂ + R₃ + …

Equivalent Resistance — Parallel

Resistors in parallel share the same voltage. The total current is the sum of the branch currents, so the reciprocals of the resistances add:

Parallel Combination V_total = V₁ = V₂ = …    (same voltage across all)
I_total = I₁ + I₂ + …
1 / R_parallel = 1/R₁ + 1/R₂ + 1/R₃ + …
For two resistors:   R_parallel = (R₁ · R₂) / (R₁ + R₂)

Kirchhoff's Laws

Voltage Law (Loop Rule)

Around any closed loop, the sum of the voltage rises (sources) equals the sum of the voltage drops (resistors). Energy is conserved — a charge returning to its starting point has gained and lost equal amounts of energy.

Current Law (Junction Rule)

At any junction, the total current flowing in equals the total current flowing out. Charge is conserved — charge cannot build up or disappear at a single point in a steady-state circuit.

Comparing Calculated and Experimental Values

In this lab you will always compute a resistance or voltage using the formula, then compare it to the experimental value you get from reading the simulated ammeter or voltmeter. Because the circuit is ideal (no meter loading, no wire resistance), the two values should agree exactly. If they disagree, re-check which formula you applied and whether you converted mA to A before dividing.

Comparison rule R_calculated = R₁ + R₂ (series) or R₁R₂/(R₁+R₂) (parallel)
R_experimental = V₀ / I_total
Both values should be equal for an ideal circuit.

Instructions — Running the Virtual Experiment

Section I — Verifying Ohm's Law

1
Select a resistor from the dropdown menu in the Section I simulation panel. The circuit shows a battery in series with the chosen resistor and an ammeter.
2
Set the battery voltage to 3 V using the slider. Read the current shown on the ammeter and record the value in the Section I data table.
3
Repeat at 6 V, 9 V, and 12 V. For each voltage, record the current. Calculate R = V / I for each row and check that all four values agree.
4
Repeat for every resistor in the dropdown. Each resistor should give a consistent R value across all four voltages, and that value should match its nominal (labelled) resistance.
5
Plot V vs I for one resistor in your lab notebook. The points should lie on a straight line through the origin; the slope of the line is the resistance.

Section II — Series and Parallel Networks

1
Select two resistors (R₁ and R₂) in the Section II panel and choose "Series" or "Parallel" from the configuration buttons.
2
Set the battery to 9 V. The simulation shows the total current from the battery, the voltage across each resistor (series) or the current through each resistor (parallel), and the calculated equivalent resistance.
3
Verify Kirchhoff's laws: for series, confirm V₁ + V₂ = V_battery; for parallel, confirm I₁ + I₂ = I_total. Record all measurements in the Section II data table.
4
Calculate the predicted equivalent resistance using R_series = R₁ + R₂ or R_parallel = R₁R₂/(R₁+R₂), and the expected voltage drop across R₁ (series: V₁ = I·R₁; parallel: V₁ = V₀). Enter these in the data table. Click Calculate Section II to see the experimental values and whether they match your calculated ones.
5
Try several pairs. Notice that R_series is always larger than either individual resistor, while R_parallel is always smaller than either. Answer the team questions and complete your lab report.

Simulation — Circuit Builder

Voltage, Resistance and Current Virtual Lab | Select a section to begin
9.0 V
Voltage
9.00
volts
Resistance
470
ohms
Current
19.15
mA
Power
0.172
W
9.0 V
R_eq
690
ohms
I_total
13.04
mA
V₁ / I₁
2.87 V
across R₁
V₂ / I₂
6.13 V
across R₂

Section I Data Table — Ohm's Law

For each resistor, calculate the current expected from Ohm's Law at each voltage (I = V / R, in mA) and type it into the "I calc" columns. Then read the actual current from the ammeter in the simulator and type it into "I exp" for each voltage. Click Calculate Section I to see a ✓/✗ match mark in each experimental cell.

Nominal R (Ω) V = 3 V V = 6 V V = 9 V V = 12 V
I calc (mA)I exp (mA) I calc (mA)I exp (mA) I calc (mA)I exp (mA) I calc (mA)I exp (mA)
100
220
470
1000
2200

Section II Data Table — Series & Parallel (V₀ = 9 V)

Enter the total current you read from the ammeter. Then type your calculated R_eq (series: R₁+R₂; parallel: R₁R₂/(R₁+R₂)), V₁ across R₁, and V₂ across R₂. In series the same current flows through both, so V₁ = I·R₁ and V₂ = I·R₂. In parallel both resistors share the full battery voltage, so V₁ = V₂ = V₀. Click Calculate Section II to reveal the experimental values and the match check.

Config R₁ (Ω) R₂ (Ω) I_total (mA) R_eq calc (Ω) R_eq exp (Ω) V₁ calc (V) V₁ exp (V) V₂ calc (V) V₂ exp (V) Match?
Series 100220
Series 220470
Series 4701000
Parallel100220
Parallel220470
Parallel4701000

Team Questions

Question 1. A 470 Ω resistor is connected to a 9 V battery. Calculate the current through the resistor in milliamperes (mA).
Question 2. Two resistors, R₁ = 220 Ω and R₂ = 470 Ω, are connected in series across a 9 V battery. Calculate the equivalent resistance in ohms.
Question 3. The same two resistors from Question 2 are now connected in parallel across the 9 V battery. Calculate the equivalent resistance in ohms (to 1 decimal place).
Question 4. In a series circuit with R₁ = 220 Ω and R₂ = 470 Ω across 9 V, what is the voltage drop across R₂? Include units (V).

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Voltage, Resistance and Current: Ohm's Law and Resistor Networks

Physics 171 | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To verify Ohm's Law (V = IR) for five resistors by measuring the current at four different voltages and checking that V/I is constant for each resistor. The lab then tests the rules for equivalent resistance in series (R_eq = R₁ + R₂) and parallel (R_eq = R₁R₂ / (R₁ + R₂)) combinations, and confirms Kirchhoff's voltage and current laws for each configuration.

Theory

Ohm's Law states that the current through an ohmic resistor is directly proportional to the voltage across it:

V = I · R

A graph of V versus I for an ohmic resistor is a straight line through the origin with slope equal to the resistance. The electrical power dissipated by the resistor is P = VI = I²R = V²/R.

For resistors in series, the same current flows through each resistor and the individual voltage drops add to the source voltage (Kirchhoff's voltage law):

R_series = R₁ + R₂      V_total = V₁ + V₂

For resistors in parallel, the same voltage appears across each resistor and the branch currents add to the total current (Kirchhoff's current law):

1 / R_parallel = 1/R₁ + 1/R₂      I_total = I₁ + I₂

Because the circuit in this simulation is ideal, the calculated value of each quantity (R_eq, V₁, I_total) should match the experimental value read from the ammeter or voltmeter exactly. Any disagreement means the wrong formula was applied, the wrong configuration was assumed (series vs parallel), or units were not converted correctly.

Calculations — Sample: R = 470 Ω at V = 9 V (Section I)

Current from Ohm's Law: I = V / R = 9.00 V / 470 Ω = 0.01915 A = 19.15 mA

Power dissipated: P = VI = (9.00)(0.01915) = 0.1723 W

Measured resistance from slope of V-vs-I graph: R = 9.00 V / 0.01915 A = 470.0 Ω (matches nominal 470 Ω)

Series sample — R₁ = 220 Ω, R₂ = 470 Ω, V₀ = 9 V:

R_series = 220 + 470 = 690 Ω
I_total = 9.00 / 690 = 0.01304 A = 13.04 mA
V₁ = I × R₁ = (0.01304)(220) = 2.87 V
V₂ = I × R₂ = (0.01304)(470) = 6.13 V
Check: V₁ + V₂ = 2.87 + 6.13 = 9.00 V ✓ (Kirchhoff's voltage law)

Parallel sample — R₁ = 220 Ω, R₂ = 470 Ω, V₀ = 9 V:

R_parallel = (220 × 470) / (220 + 470) = 103 400 / 690 = 149.86 Ω
I_total = 9.00 / 149.86 = 0.06006 A = 60.06 mA
I₁ = V / R₁ = 9.00 / 220 = 0.04091 A = 40.91 mA
I₂ = V / R₂ = 9.00 / 470 = 0.01915 A = 19.15 mA
Check: I₁ + I₂ = 40.91 + 19.15 = 60.06 mA ✓ (Kirchhoff's current law)

Results Table

Section I — Ohm's Law (currents in mA; calculated from I = V/R, experimental read from the ammeter)

Nominal R (Ω)V = 3 VV = 6 VV = 9 VV = 12 V
I calcI expI calcI expI calcI expI calcI exp
100 30.0030.00 ✓60.0060.00 ✓90.0090.00 ✓120.00120.00 ✓
220 13.6413.64 ✓27.2727.27 ✓40.9140.91 ✓54.55 54.55 ✓
470 6.38 6.38 ✓ 12.7712.77 ✓19.1519.15 ✓25.53 25.53 ✓
1000 3.00 3.00 ✓ 6.00 6.00 ✓ 9.00 9.00 ✓ 12.00 12.00 ✓
2200 1.36 1.36 ✓ 2.73 2.73 ✓ 4.09 4.09 ✓ 5.45 5.45 ✓

Section II — Series and Parallel Combinations (V₀ = 9 V)

ConfigR₁ (Ω)R₂ (Ω)I_total (mA)R_eq calc (Ω)R_eq exp (Ω)V₁ calc (V)V₁ exp (V)V₂ calc (V)V₂ exp (V)Match?
Series 100220 28.13 320.0 320.0 2.81 2.81 6.19 6.19 R ✓ · V₁ ✓ · V₂ ✓
Series 220470 13.04 690.0 690.0 2.87 2.87 6.13 6.13 R ✓ · V₁ ✓ · V₂ ✓
Series 4701000 6.12 1470.0 1470.0 2.88 2.88 6.12 6.12 R ✓ · V₁ ✓ · V₂ ✓
Parallel100220 130.91 68.75 68.75 9.00 9.00 9.00 9.00 R ✓ · V₁ ✓ · V₂ ✓
Parallel220470 60.06 149.86 149.86 9.00 9.00 9.00 9.00 R ✓ · V₁ ✓ · V₂ ✓
Parallel4701000 28.15 319.73 319.73 9.00 9.00 9.00 9.00 R ✓ · V₁ ✓ · V₂ ✓

Note: in series, V₁ + V₂ = V₀ (Kirchhoff). In parallel, V₁ = V₂ = V₀ because the resistors share the full battery voltage.

Discussion

In Section I the measured current was perfectly proportional to the applied voltage for every resistor, confirming Ohm's Law. Doubling the voltage from 3 V to 6 V exactly doubled the current; tripling the voltage to 9 V tripled it; quadrupling to 12 V quadrupled it. V / I was constant at every row within each resistor and matched the nominal resistance to three significant figures, so the V-vs-I graph for each resistor was a straight line through the origin with the correct slope. This confirms that the carbon-film resistors used in the simulation are ohmic over the voltage range tested.

In Section II the series combinations gave equivalent resistances equal to the sum of the individual resistances, and the parallel combinations gave the expected (R₁·R₂)/(R₁+R₂) values. The total current measured by the ammeter agreed with V₀/R_eq for every pair. For the representative case R₁ = 220 Ω and R₂ = 470 Ω in series, the individual voltage drops summed to exactly the battery voltage (2.87 V + 6.13 V = 9.00 V), confirming Kirchhoff's voltage law. For the same pair in parallel, the individual branch currents summed to the total current (40.91 mA + 19.15 mA = 60.06 mA), confirming Kirchhoff's current law.

The key experimental pattern is that series combinations always give a larger equivalent resistance than either component, while parallel combinations always give a smaller equivalent resistance than either component. This is why adding resistors in series reduces the current (the battery has to push charge through more resistance), while adding resistors in parallel increases the total current (the battery has additional paths through which charge can flow).

Conclusion

The experiment successfully verified Ohm's Law and the rules for combining resistors. Every resistor produced a constant V/I ratio matching its nominal resistance, so the simulated resistors are ohmic. The series rule R_eq = R₁ + R₂ and the parallel rule R_eq = R₁R₂/(R₁+R₂) were confirmed for three independent pairs in each configuration. Kirchhoff's voltage law held exactly in every series circuit and Kirchhoff's current law held exactly in every parallel circuit. Together these results confirm that the equivalent-resistance rules are direct consequences of the conservation of energy (KVL) and the conservation of charge (KCL) applied to resistor networks.

Practice Questions

Show all work and include units in your answers.

Question 1
A resistor connected to a 12 V battery draws a current of 8.0 mA. What is its resistance? What current would flow if the same resistor were connected to a 5.0 V battery?
Hint: Use R = V / I, then I = V / R with the new voltage.
Question 2
Three resistors — 100 Ω, 330 Ω, and 680 Ω — are connected in series across a 12 V battery. Find the equivalent resistance, the total current, and the voltage drop across each resistor.
Hint: R_s = R₁ + R₂ + R₃. Same current I = V/R_s flows through all. V_k = I R_k for each.
Question 3
Two resistors are connected in parallel across a 9.0 V battery. The measured total current is 75 mA and one of the resistors is known to be 150 Ω. Find the value of the other resistor and the current through each branch.
Hint: I₁ = V/R₁ through the known resistor. I₂ = I_total − I₁. Then R₂ = V / I₂.
Question 4
A 470 Ω resistor is connected to a 12 V battery. How much power (in watts) does the resistor dissipate? How much energy does it dissipate in 60 seconds?
Hint: P = V²/R. Energy E = P × t.
Question 5
Four identical 100 Ω resistors are connected in parallel across a battery. What is the equivalent resistance? If the battery voltage is 6 V, what is the total current drawn from the battery?
Hint: For N identical resistors in parallel, R_eq = R/N. Then I = V/R_eq.
Question 6 — Challenge
A 220 Ω resistor is placed in series with a parallel combination of a 470 Ω and a 1000 Ω resistor. The combination is connected to a 9 V battery. Find the total current from the battery and the voltage across the parallel section.
Hint: First compute R_parallel = (470·1000)/(470+1000). Then R_total = 220 + R_parallel. Then I = 9/R_total and V_parallel = I × R_parallel.