Voltage & current dividers
Two resistors in series split a voltage in proportion to their values. Tap the midpoint and you get Vout = Vin × R2 / (R1 + R2); with equal resistors, exactly half. The voltage divider is the standard way to scale something down to a level a chip can read. Sensing a 12 V battery with a 3.3 V microcontroller ADC starts with a divider.
The trap is loading. The formula assumes nothing draws current from the midpoint, but any real load acts like a resistor in parallel with R2 and shifts the ratio. A divider built from very large resistances gets pulled badly off by even a light load. The working rule: keep the divider’s resistance well below the load’s input resistance, or buffer the midpoint with an op-amp follower.
The current divider is the parallel twin. Current arriving at parallel branches splits inversely to their resistances, so more current takes the easier path. A branch with half the resistance carries twice the current. This also explains why an accidental short is so destructive: it’s a near-zero-resistance branch that steals nearly all the current.
Key points
- Voltage divider: Vout = Vin × R2 / (R1 + R2). The resistance ratio sets the split.
- A load on the midpoint acts in parallel with R2 and pulls the output down.
- Keep dividers stiff (low resistance) relative to their load, or buffer the output.
- Current in parallel branches divides inversely to resistance. The easier path carries more.
Practice
0 of 3 answered · Not startedWrong answers just let you try again, and hints are there if you want them. Answering every question first time, without hints, is what earns mastery.
- 1
A divider of two equal resistors from 10 V gives 5 V unloaded. You then attach a load comparable to the bottom resistor. What happens?
- 2
Why not simply use very large resistances everywhere in a divider to save power?
- 3
A 10 V source feeds a divider of R1 = 8 kΩ (top) and R2 = 2 kΩ (bottom). What is the unloaded output across R2?
V
