Coaching Centre
Series, parallel and Kirchhoff’s laws
Kirchhoff's junction rule (conservation of charge) and loop rule (conservation of energy), deriving the rules for resistors in series and parallel, solving mixed series–parallel circuits step by step, the voltage divider, and an introduction to circuits with two sources.
A modern car has more than a kilometre of wiring and dozens of separate circuits, yet an auto electrician can track down a fault with nothing more than a multimeter and two rules. One says charge can't pile up anywhere. The other says energy can't appear from nowhere. Gustav Kirchhoff wrote them down in 1845 as a student, and every circuit simulator in the world still runs on them.
Kirchhoff's two laws
The junction rule (current law) is conservation of charge. Charge cannot build up at a junction, so the total current flowing in equals the total current flowing out:
$$\sum I_{\text{in}} = \sum I_{\text{out}}$$
The loop rule (voltage law) is conservation of energy. A coulomb that travels round any closed loop and arrives back where it started must have lost exactly the energy it gained. So round any loop, the emfs add up to the potential drops across the components:
$$\sum \varepsilon = \sum IR \qquad\text{(equivalently, } \textstyle\sum \Delta V = 0 \text{ round a loop)}$$
Resistors in series
In series there is only one path, so by the junction rule the same current flows through every component. By the loop rule the potential differences add to the supply voltage: $V = V_1 + V_2 + V_3$. Writing each as IR and dividing by the common current I gives
$$R_{\text{series}} = R_1 + R_2 + R_3 + \dots$$
The total resistance is larger than any single resistor. The largest resistor takes the largest share of the voltage, since V ∝ R when I is the same.
Resistors in parallel
In parallel, each branch is connected directly across the same two points, so by the loop rule every branch has the same potential difference. By the junction rule the branch currents add: $I = I_1 + I_2 + I_3$. Writing each as V/R and dividing by the common V gives
$$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$
The total resistance is smaller than the smallest branch, because every extra branch is another path for charge. The smallest resistance carries the largest current. For two resistors, $R = \dfrac{R_1R_2}{R_1 + R_2}$ ("product over sum") is a quick shortcut.
| Series | Parallel | |
|---|---|---|
| Current | Same through each component | Splits between branches; $I = I_1 + I_2 + \dots$ |
| Potential difference | Shared; $V = V_1 + V_2 + \dots$ | Same across each branch |
| Total resistance | $R = R_1 + R_2 + \dots$ (larger than any one) | $1/R = 1/R_1 + 1/R_2 + \dots$ (smaller than any one) |
| Conservation law | Loop rule gives the voltage sharing | Junction rule gives the current sharing |
| If one globe blows | All go out | Others stay on |
Mixed circuits, step by step
To solve a circuit with both series and parallel parts: (1) replace each parallel group with its equivalent resistance; (2) add the series resistances to get the total; (3) find the total current from $I = \varepsilon/R_{\text{total}}$; (4) work back out, using V = IR for the series parts and the shared voltage for the parallel parts; (5) check with both of Kirchhoff's rules.
In the circuit above, the battery is 12 V, R₁ = 4.0 Ω, R₂ = 6.0 Ω and R₃ = 12 Ω. Find the current through and the potential difference across each resistor.
$R_{23} = \dfrac{(6.0)(12)}{6.0 + 12} = 4.0$ Ω.
$R_{\text{total}} = 4.0 + 4.0 = 8.0$ Ω, so $I = \dfrac{12}{8.0} = 1.5$ A, all of which flows through R₁.
$V_1 = IR_1 = (1.5)(4.0) = 6.0$ V. The parallel pair gets $V_{23} = (1.5)(4.0) = 6.0$ V. Loop check: 6.0 + 6.0 = 12 V ✓.
$I_2 = \dfrac{6.0}{6.0} = 1.0$ A and $I_3 = \dfrac{6.0}{12} = 0.50$ A. Junction check: 1.0 + 0.50 = 1.5 A ✓.
The voltage divider
Two resistors in series across a supply share its voltage in proportion to their resistances. Taking the output across R₂:
$$V_{\text{out}} = V\,\frac{R_2}{R_1 + R_2}$$
If R₂ is a thermistor, Vout falls as the temperature rises (its resistance drops), and an electronic circuit can use that signal to switch on a fan. This is how most temperature and light sensors feed information into a microcontroller.
A 9.0 V supply is connected across a 2.2 kΩ resistor in series with a thermistor. At 20 °C the thermistor's resistance is 4.7 kΩ. Find the potential difference across the thermistor.
$V_{\text{out}} = 9.0 \times \dfrac{4.7}{2.2 + 4.7} = 6.1$ V.
$I = \dfrac{9.0}{6900} = 1.30 \times 10^{-3}$ A, and $V = IR = (1.30 \times 10^{-3})(4700) = 6.1$ V ✓.
Going further: two sources
Some circuits can't be reduced to series and parallel parts, for example when there are two batteries in different branches. Then you apply Kirchhoff's laws directly: label an unknown current in each branch (guessing a direction), write a junction equation, write a loop equation for enough loops, and solve the simultaneous equations. A current that comes out negative simply flows the other way.
A 12 V battery with R₁ = 2.0 Ω is in the left branch, a 6.0 V battery with R₂ = 4.0 Ω in the right branch, and R₃ = 6.0 Ω in the middle branch, all joined at the top and bottom. Both batteries push current upwards in their own branches. Find the currents.
Let I₁ and I₂ flow up the side branches and I₃ down the middle: $I_1 + I_2 = I_3$.
Left loop: $12 = 2.0I_1 + 6.0I_3$. Right loop: $6.0 = 4.0I_2 + 6.0I_3$.
Substituting $I_3 = I_1 + I_2$: $12 = 8.0I_1 + 6.0I_2$ and $6.0 = 6.0I_1 + 10I_2$. Solving: $I_1 = 1.9$ A, $I_2 = -0.55$ A, $I_3 = 1.4$ A.
I₂ is negative, so 0.55 A actually flows down through the 6.0 V battery: the 12 V battery is charging it. Check the left loop: 2.0(1.91) + 6.0(1.36) = 3.8 + 8.2 = 12 V ✓.
- The default mixed circuit (12 V; 4, 6 and 12 Ω) matches the worked example: Rtotal = 8.0 Ω, I = 1.5 A, I₂ = 1.0 A, I₃ = 0.50 A, and 6.0 V across each part. Check both of Kirchhoff's laws from the readouts.
- Change R₃ to 6 Ω. Predict the total current before you look (it becomes 1.7 A, with 0.86 A in each branch).
- Make R₁ much larger than the parallel pair. Where does most of the battery's voltage go now?
- Two batteries (12 V, 6 V; 2, 4 and 6 Ω): I₂ comes out −0.55 A, so the smaller battery is being charged. Raise ε₂ until I₂ becomes positive. At what value is I₂ exactly zero, and why does that happen when V across R₃ equals ε₂?
Test yourself
Resistors of 3.0 Ω, 6.0 Ω and 9.0 Ω are connected in parallel. The total resistance is
1/R = 1/3 + 1/6 + 1/9 = 0.611, so R = 1.6 Ω, smaller than the smallest resistor.
Kirchhoff's junction rule is a consequence of the conservation of
Charge can't build up at a junction, so the current in equals the current out.
A 10 Ω and a 30 Ω resistor are in series across a 24 V supply. The potential difference across the 30 Ω resistor is
I = 24/40 = 0.60 A, so V = (0.60)(30) = 18 V. Or by proportion: 24 × 30/40.
A third identical globe is added in parallel to two identical globes already in parallel across a battery. The current from the battery
Each globe has the full battery voltage and draws the same current, so the total goes from 2 to 3 globe-currents: an increase of 50%.
In the voltage divider on this page, R₂ is replaced by a thermistor (NTC). As the thermistor warms up, Vout
An NTC thermistor's resistance falls when warm, so its share R₂/(R₁ + R₂) of the voltage falls.
Label the diagrams of Kirchhoff's rules.
Sort the statements for resistors.
Kirchhoff's junction rule states that the current entering a junction equals the current it, which follows from conservation of . The loop rule follows from conservation of : around any closed loop the sum of the emfs equals the sum of the potential . Components in always have the same potential difference across them.
A 9.0 V battery is connected to a 3.0 Ω resistor in series with a parallel combination of a 4.0 Ω resistor and a 12 Ω resistor. (a) Calculate the current through each resistor. (b) Show that your answers are consistent with Kirchhoff's two laws, naming the conservation law each one represents.
(a) Parallel pair: R = (4.0 × 12)/(4.0 + 12) = 3.0 Ω. Total R = 3.0 + 3.0 = 6.0 Ω. Current from the battery I = 9.0/6.0 = 1.5 A, which all flows through the 3.0 Ω resistor. Potential difference across the parallel pair = (1.5)(3.0) = 4.5 V. Current through the 4.0 Ω resistor = 4.5/4.0 = 1.125 A ≈ 1.1 A; through the 12 Ω resistor = 4.5/12 = 0.375 A ≈ 0.38 A.
(b) Junction rule (conservation of charge): at the junction 1.5 A enters and 1.125 + 0.375 = 1.5 A leaves, so no charge accumulates. Loop rule (conservation of energy): around the loop through the 3.0 Ω resistor and either parallel branch, the potential drops are 4.5 V + 4.5 V = 9.0 V, equal to the battery's emf, so each coulomb loses as much energy as the battery gave it.