Voltage & current dividers

YOU WILL BE ABLE TO

Derive and apply the voltage divider rule.
Derive and apply the two-branch current divider rule.
Predict and quantify the loading effect on a divider.
Recognise the potentiometer and the LDR divider as applied dividers.

Series and parallel combinations answer "what does the source see?" Dividers answer the more interesting question: "how do voltage and current split up inside the network?" These two rules are used so constantly that they deserve their own lesson.

In a series chain, every resistor carries the same current, so each resistor's voltage is proportional to its resistance — the source voltage divides in ratio. For two resistors across a source v_s, the voltage across R₂ is:

v_2 = v_s \, \frac{R_2}{R_1 + R_2}

voltage divider

The voltage divider: v_o = v_s · R_2/(R_1 + R_2), valid while the output is unloaded.

The dual: current arriving at two parallel branches splits in inverse proportion to resistance — more current takes the easier path. Note the numerator is the other branch's resistance:

i_1 = i_s \, \frac{R_2}{R_1 + R_2}

current divider, two branches

A current divider: the smaller resistance takes the larger share.

A load R_L lands in parallel with R_2 and drags the output down — the loading effect.

Two parts in the PCB Studio library are dividers in costume. A potentiometer is a resistor with a sliding third terminal — a voltage divider whose ratio you set by turning a knob; wired between 3.3 V and GND, its wiper sweeps smoothly from 0 to 3.3 V. An LDR divider pairs a light-dependent resistor with a fixed resistor: as light changes the LDR's resistance, the ratio changes, and the output voltage becomes a measurement of brightness. Both feed ADC pins — and the Studio's rules engine will refuse to place them anywhere else.

WORKED EXAMPLE

A plain divider

v_s = 3.3 V, R₁ = 10 kΩ on top, R₂ = 20 kΩ on the bottom; output taken across R₂.
v_o = 3.3 × 20k/(10k + 20k) = 3.3 × 2/3 = 2.2 V.

v_o = 2.2 V

WORKED EXAMPLE

The same divider, loaded

A 10 kΩ / 10 kΩ divider from 5 V should give 2.5 V.
Attach a 10 kΩ load across the output. The bottom is now 10k ∥ 10k = 5 kΩ.
v_o = 5 × 5k/(10k + 5k) = 1.67 V — the "2.5 V" output sagged by a third.

1.67 V loaded vs 2.5 V unloaded — loading is not a small effect

WORKED EXAMPLE

Current divider

12 mA arrives at a node feeding 6 Ω and 3 Ω in parallel.
The 3 Ω branch takes i = 12 × 6/(6+3) = 8 mA (the smaller resistor takes the larger share).
The 6 Ω branch takes the remaining 4 mA. Check: 8 + 4 = 12. ✓

8 mA through 3 Ω, 4 mA through 6 Ω

PRACTICE

A divider from 3.3 V uses R₁ = 4.7 kΩ (top) and R₂ = 10 kΩ (bottom). Find the unloaded output voltage.

Design a divider producing 1.65 V from 3.3 V. Give the resistor ratio, propose actual values, and explain why 1 Ω / 1 Ω would be a terrible choice even though the ratio is right.

10 mA enters the parallel pair 1 kΩ ∥ 4 kΩ. How much current takes the 1 kΩ branch?

An LDR sits on top of a divider from 3.3 V with a 10 kΩ fixed resistor on the bottom. In bright light the LDR is about 5 kΩ; in darkness about 100 kΩ. Find the output voltage in each case.