Resistance connected in series

Many real-life circuits are more complicated, with combinations of both series-connected and parallel-connected components. In this article, we’ll study such series-parallel circuits. You’ll find that the basic rules of the game are the ones that you’ve already learned for series circuits and parallel circuits, but the job of applying those rules can get pretty tricky.

The circuit in which the resistance is connected in series, the current flowing through each resistance is the same and the voltage drop across each resistance depends on its resistance value. Consider an electrical network shown in the figure.1, let three resistance R1, R2, and R3 be connected in series across a battery source of V volts. The current I flowing through R1, R2, and R3 is the same.

Resistance in series
Figure.1. Resistance in series

By applying Ohm’s law to the above circuit,

    \[ V_1 = 〖IR〗_1; V_2 = 〖IR〗_2; V_3 = 〖IR〗_3         ------------------- (1)  \]

By applying Kirchhoff’s Voltage Law to the above circuit,

i.e. supply voltage = sum of the voltage drop across the elements

    \[V = V_{1}+ V_{2} + V_{3}                               -----------------(2)\]

Hence, the equivalent resistance

    \[R_{T} = R_{1}+R_{2}+R_{3} ------------------(3)\]

Multiplying I on both sides of equation 3. We get

    \[I^2 R_{T}= I^2 R_{1}+I^2 R_{2}+I^2 R_{3}--------------------------------------(4)\]

Summary

Problem.1

Determine the voltage drop and power dissipated in each resistor of the circuit shown below

Problem.1- Resistance in series
Figure.2. Problem.1- Resistance in series

Step.1: Find the equivalent resistance

    \[R_{T} = R_{1}+R_{2}+R_{3} = 1 + 2 + 3 = 6 Ω\]

Total current delivered by the d.c source

    \[I = \frac{V}{R_{T}} = \frac{6}{6} =1 Amp\]

Voltage drop across 1 Ω =

    \[IR_{1} ={1}\times{1} = 1 Volt\]

Voltage drop across 2 Ω =

    \[IR_{2} ={1}\times{2} = 2 Volts\]

Voltage drop across 3 Ω =

    \[IR_{3} ={1}\times{3} = 3 Volts\]

Total voltage drop = 1+2+3 = 6 V = Supply voltage

Step.2: Power dissipated in each resistor

Power dissipated in 1 Ω  =

    \[I^{2}{R}_{1} = 1^{2}\times{1} = 1 Watt\]

Power dissipated in 2 Ω  =

    \[I^{2}{R}_{2} = 1^{2}\times{2} = 2 Watts\]

Power dissipated in 3 Ω =

    \[I^{2}{R}_{3} = 1^{2}\times{3} = 3 Watts\]

Total Power dissipated = 1+2+3 =6 watts = Power delivered by the source = V×I = 6×1 =6 V

Resistance connected in parallel

If three resistances connected in parallel, the voltage drop across each resistor are same and the current through the resistors are different. The total current supplied by the system is equal to sum of voltage drop across the resistor.

                   

Figure.3. shows three resistors are connected in parallel across a D.C source of voltage V.

Applying Ohm’s law to the network, the current through each resistor is,

    \[I_{1}= \frac{V}{R}_{1}\]

    \[I_{2}= \frac{V}{R}_{2}\]

    \[I_{3}= \frac{V}{R}_{3}\]

Applying KVL to the network,

    \[I= I_{1}+I_{2}+I_{3}\]

    \[I=\frac{V}{R_{1}} +\frac{V}{R_{2}}+\frac{V}{R_{3}}]\]

    \[\frac{V}{R_{T}}=\frac{V}{R_{1}} +\frac{V}{R_{2}}+\frac{V}{R_{3}}\]

Hence,

    \[\frac{1}{R_{T}}=\frac{1}{R_{1}} +\frac{1}{R_{2}}+\frac{1}{R_{3}}=0.5455\]

Resistance connected in parallel
Figure.3. Resistance connected in parallel

Problem.2

Figure.3 shows a sample network in which three resistors are connected in parallel across a battery of 6 V. Determine,

  1. the equivalent resistance
  2. the total current supplied by the source
  3. the power consumed by each resistor
Problem.2. Resistance in parallel
Figure.4. Problem.2. Resistance in parallel

Solution,

Step.1: Find the equivalent resistance

    \[\frac{1}{R_{T}}=\frac{1}{R_{1}} +\frac{1}{R_{2}}+\frac{1}{R_{3}}=\frac{1}{1} +\frac{1}{2}+\frac{1}{3}\]

Step.2 : Find the total current supplied by the source

    \[I=\frac{V}{R_{T}}=\frac{6}{0.5455}=11 Amps\]

Step.3: Find the current flowing through each resistor

    \[I_{1}= \frac{V}{R_{1}} =\frac{6}{1}=6 Amps\]

    \[I_{2}= \frac{V}{R_{2}} =\frac{6}{2}=3 Amps\]


    \[I_{3}= \frac{V}{R_{3}} =\frac{6}{3}=2 Amps\]

Hence,

Current through each resistor = Current supplied by the source

Step.3. Find the power consumed by each resistor

    \[P_{1}= \frac{V^{2} }{R_{1}} =\frac{6^{2}}{1}=36W\]

    \[P_{2}= \frac{V^{2} }{R_{2}} =\frac{6^{2}}{2}=18W\]


    \[P_{3}= \frac{V^{2} }{R_{3}} =\frac{6^{2}}{3}=12W\]

Sum of the power consumed by each resistor = 36 + 18 + 12 = 66 Watts

Power deviled by the source = V × I = 6 × 11 = 66 = Sum of the power consumed by each resistor.

I

Series and Parallel network

n a series-parallel circuit, the components are connected in series-connected and parallel-connected combinations. The below circuit shows an example of a series-parallel combination circuit. In this circuit, the Resistors R2, R3, and R4 are connected in series. This combination is in parallel with resistor R1. The series combination (i.e R2, R3, R4) and the parallel resistor R1 are connected to a common point called a node or Junction at which the current flowing to two paths. One is through the series combination and the other through parallel resistor R1. The amount of current division is based on the value of resistance in each path. 

Notation for Series-Parallel Relationships

Figure.5 shows the Series-Parallel network, in which three resistors such as 1ohm, 2 ohm and 3 ohm connected in parallel, and this combination is connected in series with a 6-ohm resistor. To solve the given network the following steps are to be followed,

Series - Parallel network
Figure.5. Series – Parallel network

Step.1. Reduce the parallel network by finding the equivalent resistance of parallel combination, (Ref. Figure.6)

    \[\frac{1}{R_{T}}=\frac{1}{R_{1}} +\frac{1}{R_{2}}+\frac{1}{R_{3}}=\frac{1}{1} +\frac{1}{2}+\frac{1}{3}=0.5455\]

The reduction of parallel network and the equivalent circuit is shown in Figure.5. With reference to figure.5, the reduced network has two resistors 0.5455Ω and 6Ω are connected in series. Now the problem treated similar to the resistance connected in series network.

Reduction of Series and Parallel network
Figure.6. Reduction of Series and Parallel network

Step.2. Find the equivalent resistance of the reduced network

    \[R_{T} = R_{1}+R_{2} = 0.5455 + 6 = 6.5455 Ω\]

Step.3. Find the current supplied by the source

    \[I= \frac{V}{R_{T}} = \frac{6}{6.5455} = 0.9167 Amp\]

Step.4. Find the voltage drop across each resistor

Voltage drop across 6Ω resistor = I×6Ω=0.9167×6=5.5002 V

Voltage drop across 0.5455Ω resistor = Source voltage – Voltage drop across 6Ω resistor =6V – 5.5002V = 0.5V

Step.5. Find the current through each resistor

I = V0.5455Ω /R = 0.5/1 = 0.5 Amps

I = V0.5455Ω /R = 0.5/2 = 0.25 Amps

I = V0.5455Ω /R = 0.5/1 = 0.1667 Amps

Important notation

the readers must remember the following notations while analyzing the series- parallel circuits

As you know, the three forms of Ohm’s law are:

 
Power in a Series-Parallel Circuit

The following fomulas that can be used to find the power dissipated in a resistor in a series-parallel circuit,

P = V × I

P = I2 × R

P = V2 ÷ R

where,

Total Circuit Power
Voltage-Divider Rule in Series-Parallel Circuits

The Current-Divider Rule in Series-Parallel
Circuits
Examples


Ground (or Common) Symbol
Bubble Symbol for Voltage Sources

Unit 9 Review

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