## Analysing Electric Networks

**Analysing an electric circuit** means to find the voltage and current at every point
in the circuit. A **basic electric circuit** is one where this can be done using only the
formulas for resistors in series and parallel. But there are many circuits
such as the one shown to the right where this cannot be done. Then we must use
loop analysis or nodal analysis to write down a system of equations describing the circuit.
Then we can use Gaussian elimination to solve this system of equations.

Loop analysis finds the currents directly and nodal analysis find the voltages directly. Which method is simpler depends on the given circuit. Knowing the branch currents, the nodal voltages can easily be calculated, and knowing the nodal voltages, the branch currents can easily be calculated.

## Basic Electric Circuits

Here is an electric circuit diagram:Notice these symbols:

- The symbol for a
**voltage source**(this could be for example a battery). Voltage sources generate a force or "*pressure*" which causes electric currents to flow in the circuit. The voltage source shown is rated at 10 volts (abbreviation V). This means that the voltage or "pressure" is 10 volts higher at the plus (+) side of the battery than at the minus (−) side of the battery. As a result, electric charges are given a force in the upward direction (from the − side to the + side).

- The symbol for a
**resistor**. Resistors are devices that impede the flow of electric current. The resistor at the lower right has a resistance of 55 ohms (abbrev ).

- The symbol for a
**wire**. A wire is assumed to have no resistance.

### Principal Nodes or Junctions and Branches

In this diagram we define:**principal nodes or junctions**- These are points where 3 or more wires meet. This circuit contains 4 of them, denoted N1, ..., N4. Notice the symbol for a principal node.

**branches**- A branch is any path in the circuit that has a node at each end and contains at least one voltage source or resistor but contains no other nodes. This circuit contains 6 branches, denoted B1, ..., B6.

(**Note:**If branch B4 did not contain a resistor then it could be deleted and nodes N2 and N3 could be considered one and the same node.)

### Electric Current

Electric charge flowing in a branch in a circuit is analogous to water flowing in a pipe. The rate of flow of charge is called the**current**. It is measured in coulombs/second or

**amperes**(abbreviation A) just as the flow rate of water is measured in liters/second.

Water is incompressible, which means that if 1 liter of water enters one end of a length of pipe then 1 liter must exit from the other end. The situation is the same with electric current. If the current is 1A at a certain point in a branch then it is 1A everywhere else in that branch.

### Kirchhoff's Current Law

An immediate consequence of this is**Kirchhoff's Current Law**. Kirchhoff's current law states that the sum of the currents flowing into a node equals the sum of the currents flowing out of the node. Here is an example:

This diagram also shows how we draw an arrow on the branch to indicate the current flowing in the branch.

### Electric Voltage

Electric current is the flow of electric charges.**Electric voltage is the force that causes this flow.**Just as a pump pushes a "plug" of water through a pipe by creating a pressure difference between its ends, so a battery pushes charge through a resistor by creating a voltage difference between the two ends of the resistor. The picture shows the analogy:

This diagram also shows how we draw an arrow beside a resistor or any other device to indicate a voltage difference between the two ends of that device. The arrow head is drawn pointing to the higher voltage end.

### Ohm's Law

We have just seen that a voltage difference between the two ends of a resistor causes a current to flow through the resistor. For many substances the voltage and current are proportional. This is expressed by the formula:This equation is called Ohm's law and any device that obeys it is called a resistor.V = I R,

*V*is the difference in voltage between the two ends of the resistor measured in volts,

*I*is the current through the resistor measured in amperes, and the proportionality constant

*R*is the resistance of the resistor measured in ohms. Given any two of these quantities, Ohm's law can be used to find the third.

### Kirchhoff's Voltage Law

Just as the water pressure drops in a garden hose the farther one moves away from the tap, so the voltage changes as one moves around a circuit away from a voltage source. Kirchhoff's Voltage Law states that:Around any closed path in an electric circuit, the sum of the voltage drops through the resistors equals the sum of the voltage rises through the voltage sources.A

**closed path**is a path through a circuit that ends where it starts.

**Problem: Resistors in Series.**Use Kirchhoff's voltage law and Ohm's law to find the value of the unknown resistor

*R*if it is known that a 2 ampere current flows in the circuit.

**Solution:**Let's follow the current as it flows clockwise around the circuit. If we start at

*A*and assume the voltage there is 0 then at

*B*the voltage must be 10 volts because the battery behaves like a pump that creates a higher pressure at the + side than the − side. At

*C*the voltage is still 10 volts but it drops going to

*D*through resistor

*R*, and drops again going to

*E*through the 2 ohm resistor. In fact it must return to 0 volts since

*A*and

*E*are at the same voltage (voltage does not change along an ideal wire that has no resistance).

Using Ohm's Law in the form

*V = I R*we find that the

*I R*(voltage) drop across the 2 resistor is (2 A) * (2 ) = 4 V. Then by Kirchhoff ' s Voltage Law the

*I R*drop across the unknown resistor is 10 V − 4 V = 6 V. Again using

*I*= 2 A, Ohm's law in the form

*R = V / I*gives

*R*= 3 .

The results are shown to the right. Notice the directions of the voltage arrows across each of the devices.

Also notice that the voltage drops across the two resistors are proportional to their resistances.

This is called the Voltage Divider Rule. This rule is useful in many situations.

Suppose that we replaced the above circuit by the one shown to the right and didn't know what was inside the "black box" but did know that the current flowing into the black box was 2 A and that the voltage across it was 10 V. Then Ohm's law, R = V / I, would tell us that the black box had a resistance of 5 . Notice that this is exactly the sum of the two resistances in the original circuit.

This is true in general: two resistors

*R*

_{1}and

*R*

_{2}in series may be replaced by a single equivalent resistor

*R*whose resistance is the sum of the two resistances:

_{eq}R=_{eq}R_{1}+R_{2}.

**Problem: Resistors in Parallel.**Find the value of the total current

*I*flowing into the parallel circuit.

_{T}**Solution:**Since there is no voltage drop along a wire, 20 V appears across each resistor. Using Ohm's Law in the form

*I = V / R*we find:

*I*

_{1}= 20 V / 5 = 4 A,

and:

*I*

_{2}= 20 V / 10 = 2 A,

and by Kirchhoff ' s Current Law

*I*= 4 A + 2 A = 6 A.

_{T}**Notes:**

Suppose that we replaced the above circuit by the one shown to the right and didn't know what was inside the "black box" but did know that the current flowing into the black box was

*I*= 6 A and that the voltage across it was 20 V. Let's call its resistance

_{T}*R*. Ohm's law in the form

_{eq}*I=V/R*says that:

But when we could see inside we had:

Comparing these two expressions for

*I*gives (after cancelling out the common factor of 20 V):

_{T}This formula is true in general: two resistors

*R*

_{1}and

*R*

_{2}in parallel may be replaced by a single equivalent resistor

*R*given by the formula:

_{eq}## Loop Analysis of Electric Circuits

In this method, we set up and solve a system of equations in which the unknowns are **loop currents**.
The currents in the various branches of the circuit are then easily determined from the loop currents.
(Click here for a tutorial on loop currents vs. branch currents.)

The steps in the loop current method are:

- Count the number of loop currents required. Call this number
*m*.

- Choose
*m*independent loop currents, call them*I*_{1},*I*_{2}, . . . ,*I*and draw them on the circuit diagram._{m}

- Write down Kirchhoff's Voltage Law for each loop.
The result, after simplification, is a system of
*n*linear equations in the*n*unknown loop currents in this form:

whereR_{11},R_{12}, . . . ,Rand_{mm}V_{1},V_{2}, . . . ,Vare constants._{m}

Alternatively, the system of equations can be gotten (already in simplified form) by using the inspection method.

- Solve the system of equations for the
*m*loop currents*I*_{1},*I*_{2}, . . . ,*I*using Gaussian elimination or some other method._{m} - Reconstruct the branch currents from the loop currents as described in the tutorial above.

**Example 1:**Find the current flowing in each branch of this circuit.

**Solution:**

- The number of loop currents required is 3.

- We will choose the loop currents shown to the right. In fact these loop currents are
**mesh currents**.

- Write down Kirchoff's Voltage Law for each loop. The result is the following system of equations:
Collecting terms this becomes:
This form for the system of equations could have been gotten immediately by using the
inspection method.

- Solving the system of equations using Gaussian elimination or some other method gives the
following currents, all measured in amperes:
*I*_{1}=0.245,*I*_{2}=0.111 and*I*_{3}=0.117

- Reconstructing the branch currents from the loop currents gives the results shown in
the picture to the right.

**Example 2:**Find the current flowing in each branch of this circuit.

**Solution:**

- The number of loop currents required is 3.

- This time we will choose the loop currents shown to the right.

- Write down Kirchoff's Voltage Law for each loop. The result is the following system of equations:
Collecting terms this becomes:
This form for the system of equations could have been gotten immediately by using the inspection method.

- Solving the system of equations using Gaussian elimination or some other method gives the
following currents, all measured in amperes:
*I*_{1}= −4.57,*I*_{2}= 13.7 and*I*_{3}= −1.05

- Reconstructing the branch currents from the loop currents gives the results shown in the
picture to the right.

## Nodal Analysis of Electric Circuits

In this method, we set up and solve a system of equations in which the unknowns are the**voltages at the principal nodes of the circuit**. From these nodal voltages the currents in the various branches of the circuit are easily determined.

The steps in the nodal analysis method are:

- Count the number of principal nodes or junctions in the circuit. Call this number
*n*. (A**principal node or junction**is a point where 3 or more branches join. We will indicate them in a circuit diagram with a red dot. Note that if a branch contains no voltage sources or loads then that entire branch can be considered to be one node.)

- Number the nodes
*N*_{1},*N*_{2}, . . . ,*N*and draw them on the circuit diagram. Call the voltages at these nodes_{n}*V*_{1},*V*_{2}, . . . ,*V*, respectively._{n}

- Choose one of the nodes to be the reference node or ground and assign it a voltage of zero.

- For each node except the reference node write down Kirchoff's Current Law in the form
*"the*. (By algebraic sum we mean that a current flowing into a node is to be considered a negative current flowing out of the node.)**algebraic sum**of the currents flowing out of a node equals zero"

For example, for the node to the right KCL yields the equation:*I*= 0_{a}+ I_{b}+ I_{c}

Express the current in each branch in terms of the nodal voltages at each end of the branch using Ohm's Law (*I*=*V*/*R*). Here are some examples:

The current downward out of node 1 depends on the voltage difference V1 − V3 and the resistance in the branch.

In this case the voltage difference across the resistance is V1 − V2**minus the voltage across the voltage source**. Thus the downward current is as shown.

In this case the voltage difference across the resistance must be 100 volts greater than the difference V1 − V2. Thus the downward current is as shown.

The result, after simplification, is a system of*m*linear equations in the*m*unknown nodal voltages (where*m*is one less than the number of nodes;*m = n*− 1). The equations are of this form:

where*G*_{11},*G*_{12}, . . . ,*G*and_{mm}*I*_{1},*I*_{2}, . . . ,*I*are constants._{m}

Alternatively, the system of equations can be gotten (already in simplified form) by using the inspection method.

- Solve the system of equations for the
*m*node voltages*V*_{1},*V*_{2}, . . . ,*V*using Gaussian elimination or some other method._{m}

**Example 1:**Use nodal analysis to find the voltage at each node of this circuit.

**Solution:**

- Note that the "pair of nodes" at the bottom is actually 1 extended node. Thus the number of
nodes is 3.

- We will number the nodes as shown to the right.

- We will choose node 2 as the reference node and assign it a voltage of zero.

- Write down Kirchoff's Current Law for each node. Call
*V*_{1}the voltage at node 1,*V*_{3}the voltage at node 3, and remember that*V*_{2}= 0. The result is the following system of equations: The first equation results from KCL applied at node 1 and the second equation results from KCL applied at node 3. Collecting terms this becomes: This form for the system of equations could have been gotten immediately by using the inspection method.

- Solving the system of equations using Gaussian elimination or some other method gives the
following voltages:
*V*_{1}=68.2 volts and*V*_{3}=27.3 volts

**Example 2:**Use nodal analysis to find the voltage at each node of this circuit.

**Solution:** Click here for solution.

**Example 3:**Use nodal analysis to find the voltage at each node of this circuit.

**Solution:** Click here for solution.