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Circuits: Kirchhoff's Laws, Networks and RC Transients · Lesson 4 of 6

Solving Networks: the Nodal Method

15 min read4 ideas5 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 4
The nodal method
Maths you need

Two ways to write the equations

A network with several loops can be solved by loops. Give every branch a current and use the junction rule to cut down the unknowns. Then write the loop rule round enough loops to fix the rest. That always works, but the unknowns and equations pile up quickly. There is a shorter road when many branches join the same two junctions.

Currents are driven by differences of potential. So choose the potentials of the junctions as the unknowns instead of the currents. Every branch between two junctions then has its current written in one line. The junction rule at each unknown junction gives one equation. This is the nodal method.

principle 7.1
The nodal method in three steps

First, choose one junction as the zero of potential, as if it were earthed. Second, name the potential of every other junction. Write the current in each branch as the potential difference along it, with any emf, over its resistance. Third, apply the junction rule at each junction of unknown potential and solve.

The nodal method wins most clearly when many branches run between the same two junctions. All of them then share one unknown, however many there are. The loop method would need one unknown for each independent loop. The nodal method is also Kirchhoff's two laws, used in a different order. Writing a branch current from potentials uses the loop rule along that branch.

Checkpoint · numericUnanswered

A junction A is joined through \(2\,\Omega\) to a point held at \(20\,\text{V}\), through \(1\,\Omega\) to a point held at \(10\,\text{V}\), and through \(1\,\Omega\) to a point held at \(-5\,\text{V}\). Nothing else touches A. What is the potential of A, in volts?

V± 2 %
Two cells and a load

Two cells feeding one load

A \(12\,\text{V}\) cell and a \(6\,\text{V}\) cell, each with an internal resistance of \(1\,\Omega\), are joined side by side, positive terminals together. A \(4\,\Omega\) load is joined across them. Call the bottom junction O and the top junction N. Earth O, so \(V_O = 0\), and call \(V_N = V\).

The nodal method on two cells and a loadThe same network of two cells and a load. The bottom junction O is earthed and called 0; the top junction N is given the unknown potential V. Each branch current is then written from V alone: (12 − V)/1 upward through the 12 volt branch, (6 − V)/1 upward through the 6 volt branch, and V/4 downward through the 4 ohm load. The junction rule at N sets the two upward currents equal to the downward one, (12 − V) + (6 − V) = V/4, which gives V = 8 volts. N: V O: 0 1 Ω 12 V 1 Ω 6 V 4 Ω (12 − V)/1 (6 − V)/1 V/4 (12 − V) + (6 − V) = V/4: V = 8 V
Fig 4.1Each branch current written from the one unknown \(V\): up through each cell branch, down through the load.

Walk up the \(12\,\text{V}\) branch from O: the potential rises by \(12\,\text{V}\) in the cell. The current up through its \(1\,\Omega\) is driven by what is left, \(I_1 = (12 - V)/1\). In the same way \(I_2 = (6 - V)/1\) up the \(6\,\text{V}\) branch. The load carries \(I_3 = V/4\) down. At N, the currents up equal the current down: \[ (12 - V) + (6 - V) = \dfrac{V}{4} \] Multiply by \(4\): \(72 - 8V = V\), so \(V = 8\,\text{V}\).

Now every current follows. \(I_1 = 12 - 8 = 4\,\text{A}\) and \(I_3 = 8/4 = 2\,\text{A}\). \(I_2 = 6 - 8 = -2\,\text{A}\). The minus sign means \(2\,\text{A}\) flows down through the \(6\,\text{V}\) branch, into its \(+\) terminal. So the \(6\,\text{V}\) cell is being charged. Check at N: \(4 - 2 = 2\,\text{A}\) goes down the load.

ExampleWhere the 12 V cell's power goes

For the two cells and the \(4\,\Omega\) load, find the power delivered by the \(12\,\text{V}\) cell and the power taken in by the \(6\,\text{V}\) cell. Find the heat in the resistors each second, and check that they balance.

Reveal one step at a time and finish the rest yourself.
Two cells and a load, solved by one nodeTwo cells, each with its internal resistance drawn as a resistor, and a load resistor R are joined between the same two junctions, N at the top and O at the bottom, which is earthed. The chips choose the cells: 12 volts and 6 volts with 1 ohm each; 10 volts with 1 ohm and 4 volts with 2 ohms; 4 volts and 2 volts with 1 ohm each. The slider sets R. The readouts give the potential of N and the three branch currents I_1 and I_2 upward through the cells and I_3 downward through R; each arrow's length shows the size of its current. With 12 and 6 volts and R = 4 ohms, V_N = 8 volts and the currents are 4, minus 2 and 2 amperes: the minus sign means the current in the 6 volt branch flows down, into the positive terminal, so that cell is being charged. Below R = 1 ohm the 6 volt cell gives current instead, and the arrow of its branch turns round. N O 1 Ω 12 V I1 1 Ω 6 V I2 R = 4.0 Ω I3 VN = 8.00 V, with O at 0 V I1 = 4.00 A, I2 = −2.00 A, I3 = 2.00 A the 6 V cell is being charged
SimTwo cells and a load, solved by one node. Choose the cells with the chips and slide the load \(R\). Watch \(V_N\), the three currents and the arrow of the middle branch.

With the first chip, slide \(R\) down. At \(R = 2\,\Omega\) the sim reads \(V_N = 7.2\,\text{V}\), with currents \(4.8\), \(-1.2\) and \(3.6\,\text{A}\). At \(R = 1\,\Omega\), \(V_N = 6\,\text{V}\) exactly, and the \(6\,\text{V}\) cell carries nothing. Below \(1\,\Omega\) the load pulls \(V_N\) under \(6\,\text{V}\), and the \(6\,\text{V}\) cell starts to give current: its arrow turns round.

Checkpoint · numericUnanswered

A \(9\,\text{V}\) cell and a \(3\,\text{V}\) cell, each of internal resistance \(1\,\Omega\), are joined with like terminals together across a \(1\,\Omega\) load. What is the potential difference across the load, in volts?

V± 2 %
A second network by nodes
Maths you need

A second network, unequal branches

Now a \(10\,\text{V}\) cell with \(1\,\Omega\) and a \(4\,\text{V}\) cell with \(2\,\Omega\) feed a \(2\,\Omega\) load. The branches differ, so each current carries its own resistance: \((10 - V)/1 + (4 - V)/2 = V/2\). Multiply by \(2\): \(20 - 2V + 4 - V = V\), so \(V = 6\,\text{V}\). The second chip of the sim above, with \(R = 2\,\Omega\), shows the same.

Then the \(10\,\text{V}\) branch carries \(10 - 6 = 4\,\text{A}\) up. The \(4\,\text{V}\) branch carries \((4 - 6)/2 = -1\,\text{A}\): \(1\,\text{A}\) flows down, into the \(4\,\text{V}\) cell's \(+\) terminal. The load carries \(6/2 = 3\,\text{A}\), and \(4 - 1 = 3\). The \(10\,\text{V}\) cell gives \(40\,\text{W}\). The \(4\,\text{V}\) cell takes in \(4\,\text{W}\). The resistors turn \(16 + 2 + 18 = 36\,\text{W}\) into heat.

CheckpointUnanswered

In the network of \(10\,\text{V}\) with \(1\,\Omega\), \(4\,\text{V}\) with \(2\,\Omega\) and a \(2\,\Omega\) load, the load is replaced by a \(0.5\,\Omega\) one. Which way does the current flow in the \(4\,\text{V}\) cell now?

Cells in parallel by the junction rule

Cells in parallel

Cells joined side by side, like terminals together, are a common case of the same network. Take a \(2\,\text{V}\) and a \(4\,\text{V}\) cell, each of \(1\,\Omega\), across a \(2\,\Omega\) resistor. At the top junction: \((2 - V) + (4 - V) = V/2\), so \(12 - 4V = V\) and \(V = 2.4\,\text{V}\). The \(2\,\text{V}\) cell carries \(2 - 2.4 = -0.4\,\text{A}\): it is being charged. The \(4\,\text{V}\) cell gives \(1.6\,\text{A}\), and \(1.2\,\text{A}\) reaches the load.

Deriving \(V = \dfrac{\mathcal E_1/r_1 + \mathcal E_2/r_2}{1/r_1 + 1/r_2}\) step 1 of 4
1

\[ I_1 = \dfrac{\mathcal E_1 - V}{r_1},\quad I_2 = \dfrac{\mathcal E_2 - V}{r_2} \]

Why. each cell branch, walked from the negative junction up through its emf and its internal resistance, as in the nodal method.

RememberTwo cells in parallel, on open circuit
\[ V = \dfrac{\mathcal E_1/r_1 + \mathcal E_2/r_2}{1/r_1 + 1/r_2} \]
Holds whentwo cells joined with like terminals together and nothing else joined across them; a cell turned round enters with \(-\mathcal E\).
Symbols\(V\) the voltage across the pair, \(\mathcal E_1, \mathcal E_2\) the emfs, \(r_1, r_2\) the internal resistances
Units\(V, \mathcal E\) in V, \(r\) in Ω

The voltage is a weighted average of the two emfs, each weighted by \(1/r\). A cell with a smaller internal resistance pulls the voltage towards its own emf. Two cells of \(6\,\text{V}\) and \(3\,\text{V}\), \(1\,\Omega\) each, give \(4.5\,\text{V}\). Even with no load, a current \((6 - 3)/2 = 1.5\,\text{A}\) circulates from the stronger cell into the weaker one. The chapter Cells, Bridges and Measuring Instruments uses this formula for combinations of cells.

Checkpoint · numericUnanswered

A \(2\,\text{V}\) cell of \(1\,\Omega\) and a \(4\,\text{V}\) cell of \(1\,\Omega\) are joined with like terminals together, with nothing else across them. What voltage does a perfect voltmeter read across the pair, in volts?

V± 2 %
Checkpoint · tap in orderUnanswered

Put the steps of the nodal method in order.

In the world
The home inverter's battery, charged and discharged

Many Indian homes keep a \(12\,\text{V}\) battery and an inverter for power cuts. While the mains is on, the inverter's charger acts as a source of slightly higher voltage joined to the battery, like terminals together. Current then enters the battery at its \(+\) terminal, the case of the \(6\,\text{V}\) cell in this lesson, and the battery stores energy. When the mains fails, the charger's voltage vanishes, the battery's current turns round, and the battery feeds the lights and fans.

Exam traps
  1. Forgetting the emf when writing a branch current.
    Walk the branch from one junction to the other. The current is the rise from the emf minus the fall to the far junction, over the branch resistance.
  2. Dropping a negative current as impossible.
    A negative current flows against its arrow. In a cell branch it means the cell is being charged.
  3. Averaging the emfs of cells in parallel without weights.
    Weight each emf by \(1/r\). Equal cells give a plain average; unequal ones do not.