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Physics 364: Circuit Example 2: Nodal Analysis Prof. Kroll

Not all circuits have configurations of components that make the previous approach to de-termining the voltages and the currents practical. An example of such a configuration isshown below:

This configuration is known as the “Wheatstone Bridge.” It is used to measure the value ofan unknown resistance (maybe we will talk about this more in another lecture). You cannotreduce this circuit to a single equivalent resistance using our formulas for combining resistorsin series and in parallel; we need a more systematic approach for determining voltages andcurrents in a circuit with multiple sources and tangled arrays of components.

The methods we will use are called “nodal” and “mesh” analysis. These methods are out-lined in a previous handout. In nodal analysis we determine node voltages, and this methodis particularly well suited for circuits with current sources. In mesh analysis we determinemesh currents, and this method is particularly well suited for circuits with voltage sources.

We will start with an example of nodal analysis. The circuit is shown below.

2k

10mA 50mA1k 2k

10k

First we have to identify the nodes, and then we need to define the reference node or“ground.” This circuit has three nodes, and we have defined the bottom node as the ref-erence node. We will use nodal analysis to find the voltages at nodes 1 and 2: V1 and V2.

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Always remember that what we are really determining is not absolute voltages but relativevoltages V1 − V3 and V2 − V3.

2k

10mA 50mA1k 2k

10kV1

V2

V3

Then we write down equations that express KCL using the node voltages. We need to definethe direction of our currents. The directions we have chosen are shown in the figure. Afterwe find V1 and V2 we will evaluate if we chose the correct sense of current flow. The nodalequation for node 1 is (the sum of the currents flowing into the node equals the sum of thecurrents flowing out of the node)

10 mA =V1 − V2

10k+

V1 − V2

2k+

V1

1k.

The nodal equation for node 2 is

V1 − V2

10k+

V1 − V2

2k=

V2

2k+ 50 mA.

These two equations can be simplified to the follow pair of equations:

16V1 − 6V2 = 1006V1 − 11V2 = 500

which have the solutions V1 = −957

V = −13.6 V and V2 = −3707

V = −52.9 V. (We havedropped the units in this set of equations. Our currents are in mA and our resistances arein kΩ so our voltages are in V.)

Now that we have our node voltages, we can determine the currents. We find that V1 > V2,so our assumed directions of current through the 10k and 2k resistors that are orientedhorizontally in our circuit are correct, and the values of those currents are 3.9mA and19.6mA, respectively. Both V1 and V2 are negative, and the terminology that we use toexpress this fact is that they are below ground. The assumed direction of currents thoughthe 1k and 2k resistors that are oriented vertically in our circuit is incorrect. The values ofthese currents are 13.6mA and 26.4mA, upwards.

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Circuit Example 3: Another Example of Nodal Analysis

For a second example of nodal analysis, consider the circuit below.

+

+

+

2k

2k

1k 1k

12V

12V6V

2k

As in the previous example, we must first identify the nodes, then we must choose a referencenode. There are five nodes, we chose the bottom node (V5) as our ground.

+

+

+

V1

V2 V4

V3

V5

I 12V

12V6V

2k

2k

2k

1k 1k

Since there are voltage supplies between node 2 and node 5 and node 4 and node 5, we knowV2 and V4:

V2 = −6 V V4 = 12 V.

Now we apply KCL at the other two nodes to determine V1 and V3. We will assume thatthe currents are flowing as indicated in setting up our equations, and we will substitute ourvalues for V2 and V4. Again we will not write units. Our resistors values are in kΩ and thecurrents are in mA. We start with node 1:

−6− V1

2+ I +

12− V1

2= 0.

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We were forced to introduce an additional unknown, current I, which is flowing in thedirection indicated through the 12V supply between node 3 and node 1. At node 3 we have

0 = I +V3 + 6

1+

V3

2+

V3 − 12

1.

We have two equations and three unknowns: V1, V3, and I. There is an additional constraint,however, between V1 and V3:

V1 = V3 + 12.

The solution to these three equations yields

V1 =78

7V V3 = −6

7V I =

57

7mA.

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Circuit Example 3: Mesh Analysis

In the previous example of nodal analysis, we were forced to introduce an unknown currentI in addition to our unknown node voltages Vi. Circuits with voltage sources are convenientto analyze with a different approach called mesh current analysis. In this method, we firstidentify the meshes in the circuit. Then we assign mesh currents Ii to each mesh. We useKVL around each mesh to set up a system of equations that we can then solve for themesh currents Ii, which are our unknowns. The same circuit that was used in the previousexample is shown below. There are four meshes in this circuit, and we have assigned fourmesh currents; the direction of these currents is indicated by the arrows.

+

+

+

I4

I3

I1

I2 2k

2k

1k 1k

12V6V

2k

12V

There are two currents flowing in components that are at the boundary of two meshes; forexample, in the upper 12V source, I1 flows down and I2 flows up. After we solve for thesetwo currents, we can check our answers using our previous nodal analysis by verifying thatI = I2 − I1. We will apply KVL to each mesh starting at the lower left-hand corner andproceeding clockwise around the mesh. The method does not require that all the meshcurrents are defined as flowing clockwise. They could all be counterclockwise, or some couldbe clockwise, and some could be counterclockwise. As always, the solution to the equationswill tell us the actual direction of flow, however, I find it less confusing to assign all thecurrents as all flowing clockwise or all flowing counterclockwise. The four mesh equationsare

−2I1 − 12− 1(I1 − I3) = 012− 2I2 − 1(I2 − I4) = 0

−6− 1(I3 − I1)− 2(I3 − I4) = 0−2(I4 − I3)− 1(I4 − I2)− 12 = 0.

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These equations can be simplified to

−3I1 +I3 = 12−3I2 +I4 = −12

I1 −3I3 +2I4 = 6+I2 +2I3 −3I4 = 12.

The solutions are

I1 = −60

7mA I2 = −3

7mA I3 = −96

7mA I4 = −93

7mA.

All the mesh currents flow counterclockwise; I2− I1 = 57/7mA, which is what we found forI when we solved for the node voltages.