F. Electricity and Magnetism · Electricity and MagnetismOhm's Law and Kirchhoff's Laws: Series and Parallel Circuits
F3.4 F2.2 F2.1 A1.8About 75 minutes
Learning goals- I can state Ohm's law and tell an ohmic resistor from a non-ohmic device using a V–I graph.
- I can state Kirchhoff's current law and voltage law and explain them using conservation of charge and energy.
- I can derive and use the equivalent-resistance equations for series and parallel circuits.
- I can find the current through and potential difference across each resistor in a series or a parallel circuit.
Words to know
Ohm's law
For many conductors at constant temperature, the potential difference is proportional to the current: V = IR.
ohmic / non-ohmic
An ohmic device has a constant resistance (a straight-line V–I graph through the origin). A non-ohmic device's resistance changes, for example a lamp filament as it heats up.
Kirchhoff's current law (KCL)
The total current entering a junction equals the total current leaving it (conservation of charge).
Kirchhoff's voltage law (KVL)
Around any closed loop, the total potential rise equals the total potential drop (conservation of energy).
junction
A point where three or more conductors meet and current can split or combine.
equivalent resistance (R_T)
The single resistance that would draw the same total current from the source as the whole combination.
Two laws run every circuit
Your home has dozens of circuits, and a laptop has thousands. Engineers analyse all of them with the same small toolkit: Ohm's law, which links V, I and R for one device, and Kirchhoff's two laws, which tell you how current and potential difference share out across the whole circuit. You used parts of this toolkit in Grade 9. Now you will see why the series and parallel rules work, so you can attack any circuit, including the mixed circuits in the next lesson.
Ohm's law
In the 1820s, the German physicist Georg Ohm found that for a metal wire at constant temperature, the current is proportional to the potential difference. Doubling V doubles I. We write
V = IR, or R = V ÷ I, or I = V ÷ R
A device that obeys this, with R constant, is called ohmic. Its graph of V against I is a straight line through the origin, and the slope is the resistance.
Many devices are non-ohmic. A lamp filament gets very hot as the current rises, and hot metal resists more, so its V–I graph curves upward. Diodes, LEDs and the human body are also non-ohmic. For them, R = V ÷ I still gives the resistance at that moment, but it changes.
Kirchhoff's current law (the junction rule)
Charge cannot be created or destroyed, and it does not pile up in a wire. So at any junction,
total current in = total current out
If 6.0 A enters a junction and one branch takes 2.0 A, the other branch must take 4.0 A.
Kirchhoff's voltage law (the loop rule)
Each coulomb gains energy in the source and loses it in the loads. When it returns to where it started, it must have the same energy it started with. So around any closed loop,
total potential rises = total potential drops
In a loop with one 12 V battery, the drops across all the loads in that loop must add to 12 V. Think of it like hiking: you can take any route around the mountain, but if you return to the parking lot, your total climb equals your total descent.
Series circuits, derived
In a series circuit there is only one path.
- By KCL, the current is the same everywhere: I_T = I₁ = I₂ = I₃.
- By KVL, the potential drops add up to the source: V_T = V₁ + V₂ + V₃.
- Divide each term by the common current I: V_T ÷ I = V₁ ÷ I + V₂ ÷ I + V₃ ÷ I, so
R_T = R₁ + R₂ + R₃ + …
Adding a resistor in series always increases the total resistance.
Parallel circuits, derived
In a parallel circuit, each branch connects directly across the source.
- By KVL (a loop through the source and any one branch), the potential difference is the same across each branch: V_T = V₁ = V₂ = V₃.
- By KCL, the branch currents add to the total: I_T = I₁ + I₂ + I₃.
- Divide each term by the common V: I_T ÷ V = I₁ ÷ V + I₂ ÷ V + I₃ ÷ V, so
1 ÷ R_T = 1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃ + …
Adding a branch in parallel always decreases the total resistance, because it gives the charge another path. R_T is always smaller than the smallest branch. For two branches only, a handy shortcut is R_T = (R₁R₂) ÷ (R₁ + R₂).
| Series | Parallel |
|---|
| Current | same through each: I_T = I₁ = I₂ | splits: I_T = I₁ + I₂ |
| Potential difference | splits: V_T = V₁ + V₂ | same across each: V_T = V₁ = V₂ |
| Equivalent resistance | R_T = R₁ + R₂ | 1/R_T = 1/R₁ + 1/R₂ |
| Which law gives it | KCL for I, KVL for V | KVL for V, KCL for I |
| If one load breaks | everything stops | other branches keep working |
Worked exampleExample 1: A series circuit
A 4.0 Ω resistor and a 2.0 Ω resistor are connected in series to a 12 V battery. Find R_T, the current, and the potential difference across each resistor. Check with KVL.
- Series: R_T = R₁ + R₂ = 4.0 + 2.0 = 6.0 Ω.
- Current (the same everywhere): I = V_T ÷ R_T = 12 ÷ 6.0 = 2.0 A.
- V₁ = IR₁ = (2.0)(4.0) = 8.0 V; V₂ = IR₂ = (2.0)(2.0) = 4.0 V.
- KVL check: 8.0 V + 4.0 V = 12 V ✓. Notice the bigger resistor takes the bigger share of the potential difference.
R_T = 6.0 Ω, I = 2.0 A, V₁ = 8.0 V, V₂ = 4.0 V.
A 6.0 Ω and a 3.0 Ω resistor in parallel across 12 V. Each branch has the full 12 V. The smaller resistance takes the larger current, and the branch currents add to the total (KCL).Worked exampleExample 2: A parallel circuit
A 6.0 Ω and a 3.0 Ω resistor are connected in parallel to a 12 V battery. Find each branch current, the total current and R_T two ways.
- Each branch has the full 12 V (parallel).
- I₁ = V ÷ R₁ = 12 ÷ 6.0 = 2.0 A; I₂ = 12 ÷ 3.0 = 4.0 A.
- KCL: I_T = 2.0 + 4.0 = 6.0 A.
- R_T = V_T ÷ I_T = 12 ÷ 6.0 = 2.0 Ω.
- Check with the formula: 1/R_T = 1/6.0 + 1/3.0 = 1/6 + 2/6 = 3/6, so R_T = 6 ÷ 3 = 2.0 Ω ✓. It is smaller than 3.0 Ω, the smallest branch.
I₁ = 2.0 A, I₂ = 4.0 A, I_T = 6.0 A, R_T = 2.0 Ω.
Worked exampleExample 3: Using KCL and KVL directly
(a) At a junction, 5.0 A flows in. Two wires leave carrying 1.2 A and 2.3 A, and there is one more wire. What current does it carry, and is that current entering or leaving? (b) A loop contains a 9.0 V battery and three resistors. Two of them have drops of 2.5 V and 4.0 V. Find the drop across the third.
- (a) KCL: in = out, so 5.0 = 1.2 + 2.3 + I, giving I = 1.5 A, leaving the junction.
- (b) KVL: rises = drops, so 9.0 = 2.5 + 4.0 + V₃, giving V₃ = 2.5 V.
(a) 1.5 A leaving; (b) 2.5 V.
Just for funGeorg Ohm's name became the unit of resistance. Later, engineers needed a unit for conductance (the opposite of resistance), so they spelled "ohm" backwards and called it the mho. It has since been replaced by the siemens, but you can still find "mho" on some old meters.
Hands-on activityTest the laws in a free simulation
You need: a computer or tablet with a web browser. Search for the free PhET "Circuit Construction Kit: DC" simulation (University of Colorado Boulder) and open it.
- Build a series circuit: one battery set to 12 V and two resistors set to 4 Ω and 2 Ω. Add an ammeter in series and use the voltmeter across each resistor. Compare with Example 1.
- Rebuild it as a parallel circuit with 6 Ω and 3 Ω. Put an ammeter in each branch and one in the main line. Check that the branch currents add up to the total (KCL).
- Add a third resistor in parallel. Predict first: does the total current go up or down? Then test.
- Swap one resistor for a light bulb and record V and I at three battery settings. Is the bulb ohmic in the simulation? (Some simulation bulbs are set to be ohmic; real filament bulbs are not.)
What to look for: series readings 2.0 A, 8.0 V and 4.0 V; parallel branch readings 2.0 A and 4.0 A, total 6.0 A. Adding a parallel branch increases the total current, because R_T goes down. If your measured values differ slightly from your predictions, compare them with percent difference and think about the internal resistance or meter settings in the simulation.
Your turn
1A 30 Ω resistor is connected to 9.0 V. Find the current.
I = V ÷ R = 9.0 ÷ 30 = 0.30 A.
2Find the equivalent resistance of 10 Ω, 20 Ω and 30 Ω in series.
R_T = 10 + 20 + 30 = 60 Ω.
3The resistors in the last question are connected to a 12 V source. Find the current and the potential difference across the 20 Ω resistor.
I = 12 ÷ 60 = 0.20 A. V = IR = (0.20)(20) = 4.0 V.
4Find the equivalent resistance of 20 Ω and 30 Ω in parallel.
1/R_T = 1/20 + 1/30 = 3/60 + 2/60 = 5/60, so R_T = 12 Ω. (Shortcut: (20 × 30) ÷ (20 + 30) = 600 ÷ 50 = 12 Ω.)
53.0 A enters a junction that splits into two branches. One branch carries 1.8 A. What does the other carry?
KCL: 3.0 = 1.8 + I, so I = 1.2 A.
6A 9.0 V battery runs three resistors in series. Two have potential drops of 2.5 V and 4.0 V. What is the drop across the third?
KVL: 9.0 = 2.5 + 4.0 + V₃, so V₃ = 2.5 V.
7A device gives V = 2.0, 4.0, 6.0 V at I = 0.10, 0.20, 0.30 A. Is it ohmic? What is its resistance?
R = 2.0 ÷ 0.10 = 4.0 ÷ 0.20 = 6.0 ÷ 0.30 = 20 Ω every time. The resistance is constant, so it is ohmic.
More practice: warm-up
1State Kirchhoff's current law in one sentence.
The total current into a junction equals the total current out of it.
2State Kirchhoff's voltage law in one sentence.
Around any closed loop, the total potential rises equal the total potential drops.
3In a series circuit, which quantity is the same for every resistor?
4In a parallel circuit, which quantity is the same for every branch?
The potential difference.
5Does adding another resistor in parallel make the total resistance bigger or smaller?
Smaller: it adds another path for the charge.
More practice: on your own
1Two identical 40 Ω resistors are connected in parallel. What is R_T? What would it be for five identical 40 Ω resistors in parallel?
1/R_T = 1/40 + 1/40 = 2/40, so R_T = 20 Ω. For n identical resistors R_T = R ÷ n, so five give 40 ÷ 5 = 8.0 Ω.
28.0 Ω, 12 Ω and 24 Ω resistors are in parallel across 24 V. Find each branch current, the total current and R_T.
I₁ = 24 ÷ 8.0 = 3.0 A, I₂ = 24 ÷ 12 = 2.0 A, I₃ = 24 ÷ 24 = 1.0 A. I_T = 6.0 A. R_T = 24 ÷ 6.0 = 4.0 Ω (check: 1/8 + 1/12 + 1/24 = 3/24 + 2/24 + 1/24 = 6/24 = 1/4 ✓).
3A 5.0 Ω resistor and an unknown resistor are in series with a 15 V battery. The current is 1.2 A. Find the unknown resistance.
R_T = V ÷ I = 15 ÷ 1.2 = 12.5 Ω. R = 12.5 − 5.0 = 7.5 Ω.
4A lamp has V = 1.0 V at I = 0.10 A and V = 4.0 V at I = 0.20 A. Find its resistance at each point. Is it ohmic? Explain.
R = 1.0 ÷ 0.10 = 10 Ω, then R = 4.0 ÷ 0.20 = 20 Ω. The resistance changes, so it is non-ohmic. The filament gets hotter at the larger current, and hotter metal has more resistance.
5Explain why the equivalent resistance of a parallel combination is always less than the smallest branch resistance.
The smallest branch alone would draw I = V ÷ R_small. Every other branch adds extra current on top of that at the same V, so I_T is larger, and R_T = V ÷ I_T must be smaller than R_small.
Challenge
1Use Kirchhoff's laws to derive R_T = R₁ + R₂ for series and 1/R_T = 1/R₁ + 1/R₂ for parallel.
Series: KCL gives one current I everywhere. KVL gives V_T = V₁ + V₂. Using V = IR: IR_T = IR₁ + IR₂; divide by I: R_T = R₁ + R₂. Parallel: KVL gives the same V across each branch. KCL gives I_T = I₁ + I₂. Using I = V ÷ R: V/R_T = V/R₁ + V/R₂; divide by V: 1/R_T = 1/R₁ + 1/R₂.
2A 6.0 Ω resistor is in parallel with an unknown resistor R. The combination has R_T = 4.0 Ω. Find R.
1/R = 1/R_T − 1/6.0 = 1/4 − 1/6 = 3/12 − 2/12 = 1/12, so R = 12 Ω.
3A string of 50 identical mini-lights is wired in series and plugged into 120 V. The current is 0.20 A. (a) Find the potential difference across, and resistance of, each bulb. (b) Why can one broken filament make the whole string go dark?
(a) By KVL the 50 equal drops add to 120 V, so each is 120 ÷ 50 = 2.4 V. R = 2.4 ÷ 0.20 = 12 Ω per bulb. (b) In series there is only one path. A broken filament opens the circuit, so the current everywhere becomes zero. (Many modern strings add a bypass, called a shunt, inside each bulb to prevent this.)
Quiz
1. Which law is based on conservation of charge?
2. A V–I graph for a resistor is a straight line through the origin with slope 15 V/A. What is its resistance?
3. What is the equivalent resistance of 12 Ω and 4.0 Ω in parallel?
4. Three resistors are in series with a 24 V battery. The drops across two of them are 10 V and 6.0 V. What is the drop across the third?
5. Why does a lamp filament have a curved V–I graph?
6. In a parallel circuit, a 10 Ω branch and a 20 Ω branch are connected to 10 V. Which statement is true?
Big idea: Ohm's law and Kirchhoff's laws (F3.4) are the whole toolkit for circuit analysis (F2.2). Kirchhoff's current law is conservation of charge; the voltage law is conservation of energy. The series and parallel formulas are not new rules to memorise; they follow from these two laws.
How to help: Ask your teen to explain the "potential staircase" in their own words, or to tell you why adding a branch in parallel lowers the total resistance (more paths for the charge, like opening another checkout lane).
Common difficulties:
- Forgetting the final flip in the parallel formula: 1/R_T = 0.5 means R_T = 2, not 0.5.
- Using the total voltage across a single series resistor. In series the voltage splits.
- Assuming the current is the same in parallel branches. It is the voltage that is the same.
- Calling any device with a curved V–I graph "broken". It is simply non-ohmic.