Lesson 4

Basic Laws of Electric Circuits

Nodes, Branches, Loops and Current Division

Basic Laws of Electric CircuitsNodes, Branches, and Loops:

Before going further in circuit theory, we consider thestructure of electric circuits and the names given to variousmember that make up the structure.

We define an electric circuit as a connection of electricaldevices that form one or more closed paths.

Electrical devices can include, but are not limited to,

resistors transistors transformerscapacitors logic devices light bulbsinductors switches batteries

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Basic Laws of Electric CircuitsNodes, Branches, and Loops:

A node: A node can be defined as a connection point betweentwo or more branches.

A branch: A branch is a single electrical element or device.

Figure 4.1: A circuit with 5 branches.

Figure 4.2: A circuit with 3 nodes.

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Basic Laws of Electric CircuitsNodes, Branches, and Loops:

If we start at any point in a circuit (node), proceed throughconnected electric devices back to the point (node) from which we started, without crossing a node more than one time,we form a closed-path.

A loop is a closed-path.

An independent loop is one that contains at least one elementnot contained in another loop.

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Basic Laws of Electric CircuitsNodes, Branches, and Loops:

The relationship between nodes, branches and loops can be expressed as follows:

# branches = # loops + # nodes - 1

or

B = L + N - 1 Eq. 4.1

In using the above equation, the number of loops arerestricted to be those that are independent.

In solving most of the circuits in this course, we will notneed to resort to Eq. 4.1. However, there are times when itis helpful to use this equation to check our analysis.

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Basic Laws of Electric CircuitsNodes, Branches, and Loops:

Consider the circuit shown in Figure 4.3.

+ _

Figure 4.3: A multi-loop circuit

give the number of nodesgive the number of independent loopsgive the number of branchesverify Eq. 4.15

Basic Laws of Electric CircuitsSingle Node Pair Circuits: Current division.

A single node pair circuit is shown in Figure 4.4

I

I 2 I 1

R 2 R 1

+

_

V

Figure 4.4: A circuit with a single node pair.

We would like to determine how the current divides (splits)in the circuit.

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Basic Laws of Electric CircuitsSingle Node Pair Circuits: Current division.

I

I 2 I 1

R 2 R 1V

+

_

I

R e qV

+

_

1 21 2

V VI I I

R R

eq

VIR

Therefore;

1 2

1 2 1 2

1 1 1

eq

R R

R R R R R

Eq. 4.4

Eq. 4.2 Eq. 4.3

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Basic Laws of Electric CircuitsSingle Node Pair Circuits: Current division.

From Eq. 4.4 we can write,

1 2

1 2eq

R RR

R R

Eq. 4.5

Equation 4.5 is a very important expression. In words itsays that the equivalent of two resistors in parallel equals to the product of the two resistors divided by the sum.

The equivalent resistance of two resistors in parallel is alwaysless than the smallest resistor.

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Basic Laws of Electric CircuitsSingle Node Pair Circuits: Current division.

In general, if we have N resistors in parallel as in Figure 4.5

R e q R 1 R 2 R N

Figure 4.5: Resistors in parallel.

1 2

1 1 1 1. . .

eq NR R R R Eq. 4.6

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Basic Laws of Electric CircuitsSingle Node Pair Circuits: Current division.

Back to current division: We can write from Figure 4.4;

21

1 1 1 2

eqIR IRVI

R R R R

In summary form;

21

1 2

IRI

R R

1

21 2

IRI

R R

Eq. 4.7

The above tells us how a current I divides when fed intotwo resistors in parallel. Important10

Basic Laws of Electric CircuitsSingle Node Pair Circuits: Current division.

In general, if we have N resistors in parallel and we want tofind the current in, say, the jth resistor, as shown in Figure 4.6,

Figure 4.6: General case for current division.

R e q R 1 R 2 R N R j

I

I j

eqj

j

IRI

R Eq. 4.8

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Basic Laws of Electric CircuitsCurrent Division: Example 4.1

Given the circuit of Figure 4.7. Find the currents I1 and I2

using the current division.

I1I2

4 1 2 1 0 A Fig 4.7: Circuit for Ex. 4.1.

By direct application of current division:

1

2

10(4)2.5

12 4

10(12)7.5

12 4

I A

I A

12

Basic Laws of Electric CircuitsCurrent Division: Example 4.2

Given the circuit of Figure 4.8. Find the currents I1 and I2

using the current division.

I1I2

4 1 2 1 0 A

7

Figure 4.8: Circuit for Ex. 4.2.

The 7 resistor does not change that the current toward the 4 and 12 ohm resistors in parallel is 10 A.Therefore the values of I1 and I2 are the same as in Example 4.1.

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Basic Laws of Electric CircuitsCurrent Division: Example 4.3

Find the currents I1 and I2 in the circuit of Figure 4.9 usingcurrent division. Also, find the voltage Vx

I1I2

4 1 2 2 0 V

7

+_

+

_

V x

I

Figure 4.9: Circuit for Ex. 4.3.

We first find the equivalent resistance seen by the 20 V source.

4(12)7 7 3 1012 4eqR

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Basic Laws of Electric CircuitsCurrent Division: Example 4.3

We can now find current I by,

20 202

10eq

I AR

We now find I1 and I2 directly from the current division rule:

1

2

2(4)0.5

12 4

2(12)1.5

12 4

I A

I A

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Basic Laws of Electric CircuitsCurrent Division: Example 4.3

We can find Vx from I1x12 or I2x4. In either case we get Vx = 6 V.

I1I2

4 1 2 2 0 V

7

+_

+

_

V x

I

We can also find Vx from the voltage division rule:

20(3)6

7 3xV V

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Basic Laws of Electric CircuitsCurrent Division: Example 4.4

For the circuit of Figure 4.10, find the currents I1, I2, and I3

using the current division rule.

I2I3

1 0 2 0 1 5 A 4

I1Figure 4.10: Circuit for Example 4.4.

21 3

( 15)( ) ( 15)( ) ( 15)( ), , ,

4 20 10eq eq eqR R R

I I I

1 2 3

1 1 1 1 1 1 10.25 0.05 0.1 0.4

4 20 10eqS

R R R R

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Basic Laws of Electric CircuitsCurrent Division: Example 4.4

1

2

3

15 2.59.375

4

( 15)(2.5)1.875

20

( 15)(2.5)3.75

10

I A

I A

I A

We notice that I1 + I2 + I3 = - 15 A

as expected.

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End of Lesson 4

Basic Laws of Circuits

circuits

Nodes, Branches, Loops, Current Division

Basic Laws of Electric CircuitsCurrent Division: Example 4.4

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Basic Laws of Electric CircuitsCurrent Division: Example 4.4

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