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Kirchhoff’s laws: two meshes

Adjust two sources and three resistors. Follow branch currents, voltage drops and the energy balance in two connected meshes.

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01 / Adjust the inputs

Sliders with a wide range use a logarithmic scale. When zero is allowed, it has its own position. Type a number to set an exact value.

From −48 to 48 V; the sign sets polarity.

From −48 to 48 V; the sign sets polarity.

From 1 Ω to 1 MΩ. Short circuits are not supported.

From 1 Ω to 1 MΩ. Short circuits are not supported.

From 1 Ω to 1 MΩ. Short circuits are not supported.

Try an example

A: 5 VA++I1 →I2 →R1R2R3I3 ↓E1E20 V

Arrows show positive reference directions. Each source’s upper terminal is at E1 or E2; the common return is at 0 V. A negative result means the opposite direction.

Node A voltage5 V
I1 · toward A7 mA
I2 · right2 mA
I3 · down5 mA
  • R1: 7 V · 7 mA · 49 mW
  • R2: 2 V · 2 mA · 4 mW
  • R3: 5 V · 5 mA · 25 mW

At A: 7 − 2 − 5 = 0 mA. Left mesh: 12 − 7 − 5 = 0 V. Right mesh: 5 − 2 − 3 = 0 V.

E1 supplies 84 mW; E2 absorbs 6 mW; resistors dissipate 78 mW.

Node A’s voltage determines all three currents. The shared resistor carries I1 − I2.

Use it in your classroom

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Two sources, one shared node

With E1 = 12 V, E2 = 3 V and three 1 kΩ resistors, node A is at 5 V. Through R1, (12 − 5)/1000 = 7 mA enters; through R2, (5 − 3)/1000 = 2 mA leaves; through R3, 5/1000 = 5 mA flows down. At A, current in equals current out: 7 = 2 + 5.

Follow the left mesh clockwise: the source rises 12 V, R1 drops 7 V and R3 drops 5 V. In the right mesh, also clockwise, R3 rises 5 V while R2 drops 2 V and E2 drops 3 V. Both paths return to their starting potential.

E2 absorbs 3 × 0.002 = 6 mW because current enters its positive terminal. This is a property of the ideal source, not an instruction to connect real power supplies together.

From node balance to mesh equations

I1 = (E1 − VA)/R1
I2 = (VA − E2)/R2
I3 = VA/R3
I1 − I2 − I3 = 0
VA = (E1/R1 + E2/R2)/(1/R1 + 1/R2 + 1/R3)

For clockwise mesh currents I1 and I2, the shared branch carries I3 = I1 − I2. The equivalent equations are (R1 + R3)I1 − R3 I2 = E1 and −R3 I1 + (R2 + R3)I2 = −E2. Numerical substitution and balance checks update above.

Each resistor dissipates I²R, always nonnegative. Power supplied by the sources is E1 I1 − E2 I2; their sum equals dissipation. Small numerical residuals come from rounding.

The circuit this solves

Exactly two ideal DC voltage sources and three constant resistors are modeled. Sources range from −48 to 48 V and resistors from 1 Ω to 1 MΩ. Positive resistances give a unique solution. Wire resistance, heating, current limits and transients are excluded; these results do not authorize a physical build.

Common mistakes

  • Adding I1 and I2 in R3: these reference directions require subtraction.
  • Treating a negative current as an error: actual flow is opposite to the reference direction.
  • Using E/R for a branch without subtracting node A’s potential.
  • Confusing negative voltage with negative resistor power.

Quick guide: use symmetry to check your result

  1. Start with the example and predict what happens if E2 rises from 3 to 12 V.
  2. Choose “Equal sources”: VA = 8 V; I1 = 4 mA and I2 = −4 mA. Both sources supply R3.
  3. Set both sources to zero: without applied energy, all currents vanish.
  4. Change a resistance from kΩ to Ω. The physical quantity and calculated results should stay the same.

Try it yourself

With E1 = 12 V, E2 = 0 V and three 1 kΩ resistors, what is VA?

VA = 4 V; I1 = 8 mA, I2 = I3 = 4 mA. The 0 V source is equivalent to an ideal connection to the return.

With +12 V and −12 V sources and three equal resistors, does current flow through R3?

No: VA = 0 V. I1 = I2 = 12 mA with 1 kΩ resistors; their contributions cancel in the shared branch.