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Circuits, electronics and noise
Node voltage fed through three resistors
Voltage of a node joined to three known voltages through three resistors, with no other connection.
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- ELN-04
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- v1
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- Circuits, electronics and noise
- Result
- V [V]
When three resistors join a floating node to three known voltages, Kirchhoff’s current law gives the node voltage as the average of the voltages weighted by the conductances 1/R. A smaller resistor pulls the node closer to its own voltage. The result lies between the lowest and highest source voltage.
Voltage of a node joined to three known voltages through three resistors, with no other connection. It is the conductance-weighted average of the three voltages and comes from Kirchhoff’s current law.
Calculator
- Current from source 1 into the node0.006 A
- Current from source 2 into the node0 A
- Current from source 3 into the node−0.006 A
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The node settles at 6 V, with currents of 0.006, 0 and −0.006 A flowing in from the sources, which add to zero.
- If one resistor is much smaller than the others, the node voltage is pulled to that source’s voltage.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| V1 Voltage source 1 | −50 to 50 V | −18.8 | 21.2 | 40 | +2 V | +13.3 % |
| V3 Voltage source 3 | −50 to 50 V | −14 | 26 | 40 | +2 V | +13.3 % |
| V2 Voltage source 2 | −50 to 50 V | −5.2 | 14.8 | 20 | +2 V | +6.67 % |
| R3 Resistor to source 3 | 10 to 1,000,000 Ω | 0.1478 | 9.993 | 9.846 | +10% | +3.77 % |
| R1 Resistor to source 1 | 10 to 1,000,000 Ω | 2.007 | 11.85 | 9.846 | +10% | −3.77 % |
| R2 Resistor to source 2 | 10 to 1,000,000 Ω | 6 | 6 | 1.776e-15 | +10% | 0 % |
Sweep of Voltage source 1
| Voltage source 1 (V) | Node voltage (V) |
|---|---|
| −50 | −18.8 |
| −37.5 | −13.8 |
| −25 | −8.8 |
| −12.5 | −3.8 |
| 0 | 1.2 |
| 12.5 | 6.2 |
| 25 | 11.2 |
| 37.5 | 16.2 |
| 50 | 21.2 |
Sweep one variable
Variables and how each one contributes
Node voltage
What the equation gives you. Everything else feeds it.
Voltage source 1
The voltage at the far end of resistor 1, in V. Moves the node voltage in proportion to the conductance weight of branch 1. Raising it by 10 % (to 13.2) moves the result by +8 %, from 6 to 6.48 V.
Voltage source 2
The voltage at the far end of resistor 2, in V. Moves the node voltage in proportion to the conductance weight of branch 2. Raising it by 10 % (to 6.6) moves the result by +2 %, from 6 to 6.12 V.
Voltage source 3
The voltage at the far end of resistor 3, in V. Moves the node voltage in proportion to the conductance weight of branch 3. Raising it by 10 % (to 0) moves the result by +0 %, from 6 to 6 V.
Resistor to source 1
Resistor 1 from the node to source 1, in Ω. A larger R₁ reduces the pull of source 1 on the node. Raising it by 10 % (to 1,100) moves the result by −3.77 %, from 6 to 5.774 V.
Resistor to source 2
Resistor 2 from the node to source 2, in Ω. A larger R₂ reduces the pull of source 2 on the node. Raising it by 10 % (to 2,200) moves the result by +0 %, from 6 to 6 V.
Resistor to source 3
Resistor 3 from the node to source 3, in Ω. A larger R₃ reduces the pull of source 3 on the node. Raising it by 10 % (to 1,100) moves the result by +3.77 %, from 6 to 6.226 V.
Worked example with the values above
- Conductances: 1/R₁ + 1/R₂ + 1/R₃ = 0.0025 S.
- Weighted voltages: V₁/R₁ + V₂/R₂ + V₃/R₃ = 0.015 A.
- V = 0.015 / 0.0025 = 6 V.
Assumptions and limits
- Ideal sources and linear resistors with positive values, all in base SI units.
- The circuit is exactly as described: no extra loads, and instruments that draw no current.
Common mistakes
- Forgetting a load across a divider, which lowers the output.
- Using the resistor on the wrong side of the divider or the wrong branch for the current share.
- Mixing kΩ and Ω in one calculation.
Statistics and uncertainty
Resistor values are only known to their tolerance, typically 1 % to 5 %, so a result built from them carries that scatter. Ratios such as dividers and bridges are less sensitive to a common drift than to differences between the resistors.
Where it is used
- Sensor interfaces, bias networks, bridge circuits and mixers.
- Reducing a network to simpler series and parallel parts.
Maths you need
- Ohm’s law: V = I R
- Kirchhoff’s current law: currents into a node add to zero
- Series and parallel resistors: R₁ + R₂ and R₁R₂/(R₁ + R₂)