ee1 chapter4 resistors_inparallel

Apr 12, 2023

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Contact Number

IT2001PAEngineering Essentials (1/2)

Chapter 4 - Resistors in Parallel Circuits

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IT2001PA Engineering Essentials (1/2)

Lesson Objectives

Upon completion of this topic, you should be able to: Apply Ohm’s Law to calculate voltages, currents and

resistances in a parallel circuit.

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Specific Objectives

State the characteristics of parallel-connected resistors Calculate the current flow, voltage drops across the

various resistors and the total resistance for parallel-connected resistors.

Use the current divider rule to calculate the branch current flowing in a circuit consisting of 2 resistors in parallel.

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Resistors in Parallel

Resistors are connected in parallel when the same voltage is applied across each resistor.

I

R2

R1

V

I1

I2

Two Resistors are connected in parallel

I1 & I2 are branch currents I

is the total current

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Some Examples of Parallel Circuits

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Voltage Across Resistors in Parallel

The voltage across any branch of a parallel circuit is equal to the voltage across any of the other branches in parallel.

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Voltage Across Resistors in Parallel

The voltage across each resistor is the same.

I2R2

R3I3

I

I1R1

V

V = V1 = V2 = V3

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Two Lamps Connected in Parallel

x

supplyvoltage

lamp 1lightsup

lamp 2burned out

If one lamp isburned out , the other lamp is still working.

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Example 4-1

Determine the voltage across each resistor in the figure

Solution:

The five resistors are in parallel, so the voltage drop across each one is equal to the applied source voltage.

There is no voltage drop across the fuse

V1= V2 = V3 = V4 = V5 =25V

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Current in a Parallel Circuit

I = I1 + I2

Total supply current = sum of branch currents

I2R2

I

I1 R1

V

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Example 4-2

The branch currents in the circuit of the figure are known. Determine the total current IT.

Solution

The total current IT is the sum of the two branch currents.

So the total current into point A is

IT=I1+I2=5mA + 12mA = 17mA

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Total Parallel Resistance When resistors are connected in parallel, the total resistance of

the circuit decreases. The total resistance of a parallel circuit is always less than the value of the smallest resistor.

The circuit in the figure shows a general case of ‘n’ resistors in parallel.

Applying Ohm's Law to the circuit, the total resistance is given by the equation:

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Two Unequal Resistors in Parallel

Effective resistance , R is given by

1R =

1R1

+ 1R2

=R2 + R1

R1 R2

=PRODUCT

SUM

I2R2

I1 R1

R =R1 R2

R2 + R1

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Three Resistors in Parallel

The reciprocal of the equivalent resistance

equals the sum of the reciprocals of the

branch resistances.

Parallel network Equivalent circuit

1RT

=1R1

+1R2

+1R3

RT

R2

R3

R1

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Equivalent Resistance

Equivalent or Total resistance is smaller than the lowest individual resistance.

Parallel network Equivalent circuit

R

R2

R3

R1

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Equal-Value Resistors in Parallel

Another special case of parallel circuits is the parallel connection of several resistors having the same resistance value.

Suppose there are ‘n’ number of resistors R1, R2, R3,…….Rn, all with equal resistance value of R each.

Then total resistance is given by:

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Characteristics of Parallel Circuit

The voltage across each resistor is the same.

Total current = Sum of individual current

The reciprocal of the total resistance = sum of the reciprocal of each resistance

Total resistance is smaller than the lowest individual resistance.

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Example 4-3

Find the potential difference across each resistor and the currents flowing through them.R2 =1

R3 =2

I

R1 =1

I2

I3

I1

10V

I1 = 10 / 1 = 10 AI2 = 10 / 1 = 10 AI3 = 10 / 2 = 5 A

V = 10 V potential across each resistor

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Example 4-4

Two resistors of 30 and 40 respectively are connected in a parallel. Find the total resistance.

R2

R1

1R =

1R1

+ 1R2

1R =

130

+ 140

1R =

0.0333 + 0.025

1R = 0.0583

R = 17.15

R =R2 + R1

R1 R2

R =30 + 40

30 x 40

R = 17.14

OR

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Current Divider Rule

I R R

R I

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2 1

I R R

R I

21

1 2

R1I1

I2 R2

I

R2I2

R1I1I

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Example 4-5

Find I1 and I2 .25

I2

100 I1

I = 10 A

I R R

R I

21

2 1

I1 = [ 25 / (25 +100) ] x 10 = 2 A

I R R

R I

21

1 2

I2 = [ 100 / (25 +100) ] x 10 = 8 A

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