# Analysis of resistive circuits

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Circuit_equivalence.png

A resistive circuit is a circuit containing only resistors, perfect current sources, and perfect voltage sources. This means that relationships between current and voltage are linear. The principles here also apply to phasor analysis of AC circuits.

Two electric circuits are said to be equivalent with respect to a pair of terminals if the voltages across the terminals and currents through the terminals are identical for both networks.

If [itex]V_2=V_1[itex] implies [itex]I_2=I_1[itex] for all (real) values of [itex]V_1[itex], then with respect to terminals ab and xy, circuit 1 and circuit 2 are equivalent.

 Contents

## Resistors in series and in parallel

Resistors in series: [itex]R_\mathrm{eq} = R_1 + R_2 + \,\cdots\, + R_n.[itex]

Resistors in parallel: [itex]\frac{1}{R_\mathrm{eq}} = \left( \frac{1}{R_1} \right) + \left( \frac{1}{R_2} \right) + \,\cdots\, + \left( \frac{1}{R_n} \right).[itex]

Special case: Two resistors in parallel: [itex]R_\mathrm{eq} = \left( \frac{R_1R_2}{R_1 + R_2} \right).[itex]

## Delta-wye transformation

The transformation is used to establish equivalence for networks with 3 terminals.

For equivalence, the resistance between any pair of terminals must be the same for both networks.

## Delta-to-wye transformation equations

[itex]R_1 = \left( \frac{R_aR_b}{R_a + R_b + R_c} \right)[itex]
[itex]R_2 = \left( \frac{R_bR_c}{R_a + R_b + R_c} \right)[itex]
[itex]R_3 = \left( \frac{R_cR_a}{R_a + R_b + R_c} \right)[itex]

## Wye-to-delta transformation equations

[itex]R_a = \left( \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_2} \right)[itex]
[itex]R_b = \left( \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_3} \right)[itex]
[itex]R_c = \left( \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_1} \right)[itex]

## Voltage and current division

Voltage division: Consider n resistors that are connected in series. The voltage across any resistor [itex]R_i[itex] is

[itex]V_i = R_iI = \left( \frac{R_i}{R_1 + R_2 + \cdots + R_n} \right)V[itex]

Current division: Consider n resistors that are connected in parallel. The current [itex]I_i[itex] through any resistor [itex]R_i[itex] is

[itex]I_i = \left( \frac{\left( \frac{1}{R_i} \right)}{ \left( \frac{1}{R_1} \right) + \left( \frac{1}{R_2} \right) + \,\cdots\, + \left( \frac{1}{R_n} \right)} \right)I[itex]

for [itex]i = 1,2,...,n.[itex]

### Special case: Two resistors in parallel

[itex]I_1 = \left( \frac{R_2}{R_1 + R_2} \right)I[itex]
[itex]I_2 = \left( \frac{R_1}{R_1 + R_2} \right)I[itex]

## Source transformation

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Sourcetransform.PNG
Image:Sourcetransform.PNG

If the two networks are equivalent with respect to terminals ab, then V and I must be identical for both networks. Thus

[itex]V = RI_s[itex] or [itex]I_s = \left( \frac{V_s}{R} \right)[itex]

See also: Norton's theorem, Thevenin's theorem

## Nodal analysis

### General procedure

1. Label all nodes in the circuit. Arbitrarily select any node as reference.

2. Define a voltage variable from every remaining node to the reference. These voltage variables must be defined as voltage rises with respect to the reference node.

3. Write a KCL equation for every node except the reference.

4. Solve the resulting system of equations.

## Mesh analysis

### General procedure

Mesh - a loop that does not contain an inner loop.

1. Count the number of “window panes” in the circuit. Assign a mesh current to each window pane.

2. Write a KVL equation for every mesh whose current is unknown.

3. Solve the resulting equations.

## Choice of method

Given the choice, which method should be used: nodal analysis or mesh analysis?

Nodal analysis: The number of voltage variables equals number of nodes minus one. Every voltage source connected to the reference node reduces the number of unknowns by one.

Mesh analysis: The number of current variables equals the number of meshes. Every current source in a mesh reduces the number of unknowns by one.

There is also another point to consider: mesh analysis only applies to planar circuits, i.e. circuits that can be drawn using only two dimensions. Intuitively, what this means is that the wires in the circuit diagram must not "jump over" one another if one is to apply mesh analysis. Nodal analysis, on the other hand, can be applied to both planar and non-planar circuits. Note that sometimes, a circuit that is drawn in a non-planar fashion (i.e. with wires jumping over each other) may be redrawn in planar form, although this is not always the case. Generally, most of the circuits one encounters in elementary resistive network analysis are planar in nature.

To summarize, for planar circuits, either nodal or mesh analysis may be used; generally, the method with the least unknowns to solve for is selected. For circuits that are non-planar, one can try to redraw the circuit in planar form; if this is not possible, one has no choice but to apply nodal analysis.

## AC circuits

All the techniques given above can be applied to single freqency AC circuits by using phasors represented as complex numbers for voltage and current and using complex impedance in place of resistance.

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