Voltage node method

The idea

When a circuit has more than one source, simple series–parallel reduction stops working. The node method picks one unknown, the voltage at the main junction, and writes KCL there with every current expressed as $\Delta V/R$. One equation, one unknown.

The method

1. Mark the nodes and choose a reference node (ground, 0 V). Every node voltage is measured from it.
2. Draw an assumed direction for each branch current.
3. At the principal node N, write KCL with each current as (voltage at one end − voltage at the other) ÷ R.
4. Solve for $V_N$.
5. Work out every current, voltage drop and power, then check the power balance.

Worked example

Two sources share a load. $V_A=24$ V connects to node N through $R_1=6\Omega$. $V_B=12$ V connects through $R_3=4\Omega$. A load $R_2=12\Omega$ goes from N to ground. Find $V_N$ and every current.

Answer

KCL at N (currents from the sources in = current to ground out): $\frac{24-V_N}{6}+\frac{12-V_N}{4}=\frac{V_N}{12}$ Multiply by 12: $2(24-V_N)+3(12-V_N)=V_N$ → $84=6V_N$ → $V_N=14$ V. $I_1=(24-14)/6=1.67$ A · $I_3=(12-14)/4=-0.5$ A · $I_2=14/12=1.17$ A $I_3$ is negative: current flows into the 12 V source, which is being charged. KCL check: $1.67-0.5=1.17$ ✓ Power: resistors $16.7+1.0+16.3=34$ W. Sources: $24\times 1.67-12\times 0.5=40-6=34$ W ✓

A source with negative power is absorbing energy, like a robot’s battery being charged by the main supply.

Common mistakes

• Mixing directions: if every current is written as “flowing out of N”, then $\sum =0$. If some are in and some out, then in = out. Pick one style and stick with it.
• Forgetting to multiply through by a common denominator before rearranging.

Check yourself

1. In the example, what would $V_N$ be if the 12 V source were removed (its branch disconnected)?

Now it’s a plain divider: $24\times \frac{12}{6+12}=16$ V.