Elec Class
learn electricity hands-on
Lesson · chapter 6 of 15
In series, resistances add up: Réq = R₁ + R₂. In parallel, their reciprocals add up (1/Réq = 1/R₁ + 1/R₂), and for two resistors Réq = R₁ × R₂ / (R₁ + R₂).
The result is worked out before your eyes, line by line, just like on the board.
1I'm looking for
2I enter the known values
Formula used Réq = R₁ · R₂R₁ + R₂
3The result
Equivalent resistance Réq600 Ω
{R₁} in parallel with {R₂} is equivalent to a single resistance of {Réq}.
The definition, the key points to remember and everyday examples.
In series the same current flows through every resistor: resistances add up and the voltage is shared in proportion to each (voltage divider).
In parallel all branches are at the same voltage: currents add up and the equivalent resistance is smaller than the smallest branch. A mixed network is reduced step by step. The two wires of a line are in series with the load: they cause a voltage drop 2·R_l·I.
In parallel, opening one more checkout adds a path: the line moves faster, even if the new checkout is slow, just as Réq always drops when you add a branch. In series, it's like going through two checks one after the other: the waits add up. Where it falls short: a customer chooses a checkout, whereas the current divides by itself, in inverse proportion to the resistances.
Whatever the network
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage across resistor n | Un | V |
| Resistor n | Rn | Ω |
| Current through it | In | A |
Every resistor obeys Ohm's law with its own voltage and current, in series as in parallel.
| Quantity | Symbol | Unit |
|---|---|---|
| Equivalent resistance | Réq | Ω |
| Voltage across the network | U | V |
| Total current | I | A |
Req is the single resistor that, at the same voltage U, would carry the same current I.
| Quantity | Symbol | Unit |
|---|---|---|
| Power dissipated by resistor n | Pn | W |
| Quantity | Symbol | Unit |
|---|---|---|
| Total power | P | W |
Powers always add up, in series, in parallel or in a mixed network.
In series
| Quantity | Symbol | Unit |
|---|---|---|
| Equivalent resistance | Réq | Ω |
| Quantity | Symbol | Unit |
|---|---|---|
| Current (common) | I | A |
| Total voltage (split) | U | V |
In series the current I is common to every resistor; the voltage U is split between them.
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage across resistor n | Un | V |
| Resistor n | Rn | Ω |
| Current | I | A |
| Quantity | Symbol | Unit |
|---|---|---|
| Current | I | A |
| Total voltage | U | V |
| Equivalent resistance | Réq | Ω |
| Quantity | Symbol | Unit |
|---|---|---|
| Resistor n | Rn | Ω |
| Voltage across resistor n | Un | V |
| Current | I | A |
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage on R₁ | U₁ | V |
| Total voltage | U | V |
In parallel
| Quantity | Symbol | Unit |
|---|---|---|
| Equivalent resistance | Réq | Ω |
Two resistors: Req = R₁·R₂ / (R₁ + R₂).
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage (common) | U | V |
| Total current (split) | I | A |
In parallel the voltage U is common to every branch; the current I is split between them.
| Quantity | Symbol | Unit |
|---|---|---|
| Current in branch n | In | A |
| Voltage (common) | U | V |
| Resistance of branch n | Rn | Ω |
The branch with the smallest resistance carries the largest current.
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage (common) | U | V |
| Total current | I | A |
| Equivalent resistance | Réq | Ω |
| Quantity | Symbol | Unit |
|---|---|---|
| Resistance of branch n | Rn | Ω |
| Voltage (common) | U | V |
| Current in branch n | In | A |
| Quantity | Symbol | Unit |
|---|---|---|
| Current in R₁ | I₁ | A |
| Total current | I | A |
Two branches: the current splits inversely to the resistances (I₁ gets R₂'s share).
| Quantity | Symbol | Unit |
|---|---|---|
| Equivalent resistance | Réq | Ω |
| Resistance of each branch | R | Ω |
| Number of resistors | n | - |
Mixed network
| Quantity | Symbol | Unit |
|---|---|---|
| Equivalent resistance of R₂ ∥ R₃ | R₂₃ | Ω |
| Total equivalent resistance | Réq | Ω |
Example: R₁ in series with R₂ ∥ R₃. Replace each series or parallel group by its equivalent resistance until a single resistor remains.
| Quantity | Symbol | Unit |
|---|---|---|
| Equivalent resistance of R₂ + R₃ | R₂₃ | Ω |
| Total equivalent resistance | Réq | Ω |
R₁ in parallel with the R₂ + R₃ branch: reduce the series branch first, then the parallel.
| Quantity | Symbol | Unit |
|---|---|---|
| Current delivered by the generator | I | A |
| Generator voltage | U | V |
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage on R₁ | U₁ | V |
| Voltage shared by R₂ and R₃ | U₂₃ | V |
| Current in R₂ | I₂ | A |
Walk back the other way: a series group gets the common current (U = R · I), a parallel group gets the common voltage (I = U / R). Check: U₁ + U₂₃ = U and I₂ + I₃ = I.
Power line
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage drop | ΔU | V |
| Resistance of one wire | Rl | Ω |
| Current | I | A |
Four steps to solve a problem, then an example solved line by line.
Series: all the current from one goes into the other, with no branching. Parallel: connected between the same two points.
Replace each group with its equivalent resistance, starting with the innermost one, until only one resistance is left.
I = U / Réq, with the source voltage.
Series group: U = R × I with the shared current; parallel group: I = U / R with the shared voltage. The sums must add back up to U and to I.
R₁ = 100 Ω is in series with the group R₂ = 220 Ω ∥ R₃ = 330 Ω, all under U = 12 V. Find the equivalent resistance, the total current, then the voltage and current of each resistor.
Parallel first: R₂₃ = 220 × 330 / (220 + 330) = 132 Ω. Then series: Req = 100 + 132 = 232 Ω.
I = U / Req = 12 / 232 = 51.7 mA.
U₁ = R₁ · I = 100 × 0.0517 = 5.17 V; U₂₃ = 132 × 0.0517 = 6.83 V (check: 5.17 + 6.83 = 12 V).
I₂ = U₂₃ / R₂ = 6.83 / 220 = 31.0 mA; I₃ = 6.83 / 330 = 20.7 mA (check: 31.0 + 20.7 = 51.7 mA).
Add the reciprocals, then take the reciprocal again: 1/Réq = 1/R₁ + 1/R₂ + ... For two resistors, it's faster with Réq = R₁ × R₂ / (R₁ + R₂): 100 Ω and 400 Ω give 40,000 / 500 = 80 Ω.
With the reciprocals: for 100, 200 and 400 Ω, 1/Réq = 0.01 + 0.005 + 0.0025 = 0.0175, so Réq = 1 / 0.0175 ≈ 57.1 Ω. The "product over sum" formula only works for two resistors at a time.
In series, n times the value of one: 4 resistors of 220 Ω make 880 Ω. In parallel, n times less: the same 4 resistors make 220 / 4 = 55 Ω.
Follow the current: if all the current leaving one goes into the other, with no branching between them, they are in series. If they are connected between the same two points, they are in parallel, whatever the drawing looks like.
To get a voltage smaller than the source voltage with two resistors in series: U₂ = U × R₂ / (R₁ + R₂). With 12 V, R₁ = 10 kΩ and R₂ = 5 kΩ, you get 4 V, as long as the device connected across R₂ draws little current.
R₂ = R₁ × Réq / (R₁ − Réq). To go from 1 kΩ to 750 Ω, you need R₂ = 1,000 × 750 / 250 = 3,000 Ω, so 3 kΩ in parallel.
Each branch offers an extra path to the current: under the same voltage the total current is larger, so U/I is smaller. Two equal resistors in parallel are worth half of one.
In parallel: each receives the 230 V of the mains and works independently of the others. In series, switching one lamp off would switch off all the others, and each would receive only a fraction of the voltage.
From the simplest to the most complete, each problem comes with a step-by-step solution.
Give each group three resistors (100 Ω, 220 Ω, 330 Ω) and have them find every way to combine them: students calculate each Réq, measure it with the ohmmeter, then rank the circuits from the smallest (about 57 Ω, all in parallel) to the largest (650 Ω, all in series).
Using Elec Class in class