Nodal analysis is a circuit-analysis format that combines Kirchhoff’s current- law equations with the

source transformation. Converting all voltage sources to equivalent constant-current sources allows us to standardize the way we write theKirchhoff’s current-lawequations.

For nodal analysis, we consider source currents to flow into a node. If the arrows in a circuit diagram showing that a source current actually flows out of a particular node, the current is a negative quantity for that node.

**Similarly**, we consider all resistor currents to flow out of a node. We are thus restating Kirchhoff’s current law in the form:

At any independent node, the algebraic sum of the resistor currents leaving the node equals the algebraic sum of the source currents entering the node.

The Kirchhoff’s current-law equations then have the format

$\begin{matrix} {{I}_{R1}}+{{I}_{R2}}+\cdots ={{I}_{S1}}+{{I}_{S2}}+\cdots & {} & \left( 1 \right) \\\end{matrix}$

The next step is to replace the resistor currents using I = VG.

The voltage across each resistor equals the difference between the voltages, relative to the reference node, of the nodes at each end of the resistor. We always subtract the voltage of the adjacent node from the voltage of the node for which we are writing the equation. For node 1 in

**Figure 1**, the current through R_{1} equals V_{1}G_{1}. The voltage across R_{3} is V_{1 }– V_{2}. Hence the Kirchhoff’s current- law equation for node 1 becomes

${{V}_{1}}{{G}_{1}}+\left( {{V}_{1}}-{{V}_{2}} \right){{G}_{2}}={{I}_{S1}}$

Collecting the voltage terms gives

\[\begin{matrix} \left( {{G}_{1}}+{{G}_{3}} \right){{V}_{1}}-{{G}_{3}}{{V}_{2}}={{I}_{S1}} & {} & \left( 2 \right) \\\end{matrix}\]

**Figure 1** circuit diagram for writing nodal equations

Equation 2 illustrates the standard format for writing a nodal equation. We can use determinants to solve for the voltages.

We can write **Equation 2** directly by noting that the positive voltage term on the left-hand side of the equation is the unknown voltage for the node in question multiplied by the sum of the conductance connected to that node. **From this positive term**, we must subtract a term for the node voltage at every adjacent node connected by a resistor to the node in question. This term is the adjacent node voltage multiplied by the conductance between the two nodes. Using this procedure to write the nodal equation for node 2 gives

\[\left( {{G}_{2}}+{{G}_{3}} \right){{V}_{2}}-{{G}_{3}}{{V}_{1}}=-{{I}_{S2}}\]

Nodal analysis is particularly useful for networks where a common portion of the network is fed from several sources in parallel.

**Nodal Analysis Example 1**

An automobile generator with an internal resistance of 0.20 V develops an open-circuit voltage of 16.0 V. the storage battery has an internal resistance of 0.10 V and an open-circuit voltage of 12.8 V. Both sources are connected in parallel to a 1.0-V load. Determine the load current.

**Solution**

**Step 1**

First, we draw the equivalent circuit with the voltage sources converted to constant-current sources, as shown in **Figure 2**.

**Figure 2** Circuit diagram for Example 1

The short-circuit current for the battery is

\[{{I}_{bat}}=\frac{{{E}_{bat}}}{{{R}_{bat}}}=\frac{12.8V}{0.10\Omega }=128A\]

Similarly, the short-circuit current of the generator is

\[{{I}_{gen}}=\frac{{{E}_{gen}}}{{{R}_{gen}}}=\frac{16V}{0.20\Omega }=80A\]

With a bit of mental arithmetic, we can fill in the required data on the circuit diagram of Figure 2.

**Step 2**

Next, we find the conductance of each branch:

$\begin{align} & {{G}_{gen}}=\frac{1}{{{R}_{gen}}}=\frac{1}{0.20\Omega }=5S \\ & {{G}_{bat}}=\frac{1}{{{R}_{bat}}}=\frac{1}{0.10\Omega }=10S \\ & and \\ & {{G}_{L}}=\frac{1}{{{R}_{L}}}=1S \\\end{align}$

**Step 3**

At first glance, Figure 2 appears to show six junction points, but there are only two because the same voltage appears across all parallel branches. For our nodal analysis, we can redraw the circuit in the form shown in **Figure 3**.

If we label the lower node as the reference node, the upper node is the only independent node. Consequently, we require only one **Kirchhoff’s current-law** equation for the circuit of Figure 3.

**Figure 3** Equivalent circuit for Example 1

At the single independent node, Equation 1 becomes

${{I}_{5}}+{{I}_{1}}+{{I}_{10}}={{I}_{gen}}+{{I}_{bat}}$

Or, we can go directly to the format of Equation 2,

$\begin{align} & \left( {{G}_{5}}+{{G}_{1}}+{{G}_{10}} \right)V={{I}_{gen}}+{{I}_{bat}} \\ & V=\frac{80+128}{5+1+10}=13V \\ & {{I}_{L}}={{G}_{L}}{{V}_{L}}=13V\times 1S=13A \\\end{align}$

The nodal analysis allowed us to solve Example 1 with not much more than mental arithmetic since the circuit has only a single independent node. **However**, we do need to solve simultaneous equations when the network has more than one independent node.