Voltage in Parallel Calculator

Two components are in parallel when both ends connect to the same pair of nodes. That is why the defining rule of a parallel circuit is about voltage:

$V_1 = V_2 = \cdots = V_{\text{supply}} \qquad I_{\text{total}} = I_1 + I_2 + \cdots$

Symbols and units: $V$ in volts (V), $I$ in amperes (A), $R$ in ohms (Ω).

Each branch current is found independently from Ohm's law, $I_n = V/R_n$, and the equivalent resistance comes from summing conductances:

$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

The result is always smaller than the smallest branch, because adding a path can only make it easier for current to flow.

The operating assumption: an ideal source that holds its voltage regardless of load, and wiring with no resistance of its own. Section two removes that second assumption, which is where most real-world errors live.

The mistake people make: adding parallel resistances. Adding is the series rule.

Once the wiring itself is included, the branches no longer share exactly the same voltage — the cable steals some of it. A conductor's resistance is

$R = \frac{\rho L}{A}$

with $\rho$ in ohm-metres (copper is $1.72 \times 10^{-8}$ Ω·m at 20 °C), $L$ the conductor length in metres and $A$ the cross-sectional area in square metres. Note that $1$ mm² $= 1 \times 10^{-6}$ m².

For a two-core supply the current travels out and back, so the loop length is twice the route length:

$L_{\text{loop}} = 2 \times L_{\text{route}} \qquad \Delta V = I R_{\text{loop}}$

The loop resistance also sets the prospective fault current, $I_f = V / Z_{\text{loop}}$, where $Z_{\text{loop}}$ adds the supply's own impedance to the cable's.

These are teaching calculations. Conductor size, permissible voltage drop and disconnection time for a real installation are set by the applicable wiring code — the NEC, BS 7671 or the local equivalent — using its own tables, correction factors and design current. Never install to an arithmetic result alone.

• Adding resistances in parallel — sum the reciprocals, then invert the sum once at the end.
• Using the one-way length for a cable run — the current goes and returns, so a $30$ m route is $60$ m of conductor.
• Assuming every parallel branch really sees the source voltage — with long cable runs it does not, and the far branch is starved.
• Leaving the cross-section in mm² — $2.5$ mm² is $2.5 \times 10^{-6}$ m² in an SI formula.
• Using the 20 °C resistivity for a hot conductor — a cable at $70$ °C has roughly $20\%$ more resistance, which is precisely the condition voltage-drop checks care about.
• Ignoring the source impedance in a fault calculation — the transformer and the service cable are part of the loop.
• Confusing conductor cross-section with cable diameter — the overall sheath is much larger than the copper inside it.