Series, parallel and Kirchhoff's laws

Real circuits are never one resistor and one battery. Components are connected to each other in only two basic ways, series and parallel, and every complicated network is a mixture of the two. With two conservation rules, Kirchhoff's laws, you can predict every voltage and current in any such network before you touch a wire.

Series: one path

Components are in series when they are chained end to end, so the same current must pass through all of them. There is no junction where the current could split.

Because the voltages add up along the chain and the current is shared, the total resistance is simply the sum:

R_eq = R1 + R2 + R3

Three resistors of 100 Ω, 220 Ω and 330 Ω in series give R_eq = 100 + 220 + 330 = 650 Ω. Adding more resistors in series always raises the total.

Parallel: shared voltage

Components are in parallel when both of their ends are joined together. Each one sees the same voltage, and the current splits among them. More paths make it easier for current to flow, so adding a parallel resistor always lowers the total resistance. Conductance (1/R) is what adds up:

1 / R_eq = 1 / R1 + 1 / R2 + 1 / R3

Two resistors of 100 Ω and 300 Ω in parallel:

1. 1 / R_eq = 1/100 + 1/300 = 3/300 + 1/300 = 4/300
2. R_eq = 300 / 4 = 75 Ω

For exactly two resistors there is a shortcut, "product over sum": R_eq = R1 * R2 / (R1 + R2) = 100 * 300 / 400 = 75 Ω. The result is always smaller than the smallest branch. Two equal resistors in parallel give exactly half of one.

Kirchhoff's two laws

Gustav Kirchhoff turned two conservation ideas into tools.

Kirchhoff's current law (KCL): the currents flowing into any junction equal the currents flowing out. Charge does not pile up or vanish at a node. If 6 mA arrives at a junction and one branch takes 2 mA, the other takes 4 mA.

Kirchhoff's voltage law (KVL): going around any closed loop, the voltage rises and drops add up to zero. Put differently, the sum of the drops across the components equals the voltage of the source. Energy given to a coulomb by the battery is exactly spent by the time it returns.

Everything in the next worked example is just these two laws plus Ohm's law (V = I * R).

A worked circuit

A 12 V source feeds R1 = 1 kΩ in series with a parallel pair, R2 = 2 kΩ and R3 = 2 kΩ. Find every voltage and current.

Step 1, reduce the parallel part. Two equal resistors in parallel give half: R23 = 2 kΩ / 2 = 1 kΩ.

Step 2, add the series part. R_total = R1 + R23 = 1 kΩ + 1 kΩ = 2 kΩ.

Step 3, total current. I = V / R_total = 12 V / 2 kΩ = 6 mA. This is the current leaving the source, and since R1 is in series with everything else, it is also the current through R1.

Step 4, voltage across R1. V1 = I * R1 = 6 mA * 1 kΩ = 6 V.

Step 5, voltage across the parallel pair. By KVL, 12 V - 6 V = 6 V. Check with Ohm: 6 mA * 1 kΩ = 6 V.

Step 6, branch currents. Each branch sees 6 V: I2 = 6 V / 2 kΩ = 3 mA, and I3 = 6 V / 2 kΩ = 3 mA. By KCL, 3 mA + 3 mA = 6 mA, which matches the current entering the junction.

Step 7, check the energy. The source delivers 12 V * 6 mA = 72 mW. The resistors dissipate 36 mW (R1) + 18 mW + 18 mW = 72 mW. Every milliwatt is accounted for. Always do this last check: if it fails, you made an arithmetic slip somewhere.

Dividing voltage

Two resistors in series share the supply in proportion to their resistance, because the same current flows through both. This is the voltage divider:

Vout = Vin * R2 / (R1 + R2)

Drag the sliders below. Notice that Vout depends only on the ratio of R1 to R2, not on their absolute size, and that Vout is always between 0 V and Vin.

Vin5 V

R110 kΩ

R210 kΩ

Vout = Vin × R2 / (R1 + R2) = 2.500 V

A divider only holds its voltage while nothing draws current from the output. Connect a load in parallel with R2 and the lower resistance changes the ratio. With Vin = 5 V and R1 = R2 = 10 kΩ, the unloaded output is 2.5 V. Hang a 10 kΩ load on the output: R2 || R_load = 5 kΩ, so Vout = 5 V * 5 / (10 + 5) = 1.67 V. This is why dividers feed high-impedance inputs such as an ADC pin and are not used to power things.

Why LEDs in parallel need their own resistors

Wiring two LEDs in parallel behind a single shared resistor looks economical and is a classic mistake. There are two reasons.

Reason 1, the current does not split evenly. An LED's current rises extremely steeply with voltage. Two LEDs of the same type still have slightly different forward voltages, perhaps 1.95 V and 2.05 V. With both sitting at the same node voltage, the one with the lower forward voltage takes far more than its fair share of current, runs hotter, and degrades faster. The shared resistor cannot correct this because it only sees the total.

Reason 2, a failure cascades. Take a 5 V supply, LEDs dropping 2 V, and a target of 10 mA each. A shared resistor must pass 20 mA: R = (5 - 2) / 0.020 = 150 Ω. If one LED fails open, all 20 mA now goes through the survivor, double its design current.

With an individual resistor per LED, each branch is set independently: R = (5 - 2) / 0.010 = 300 Ω per LED. A fault in one branch cannot change the current in the other, and small differences in forward voltage barely matter because the resistor, not the LED, sets the current. The supply delivers the same 20 mA in total, so the cost is one extra resistor.