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Resistors in Series and Parallel

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Resistance

Resistance is defined as the measure of opposition to the motion of electrons due to their continuous collisions, with the atoms of the conductor.

The unit of resistance is ohm.

Let’s consider the conductor of cross-sectional area A and of length L as given below:

Conductor

The resistance of this conductor is

1.Directly proportional to its length i.e. the resistance increases with increase in length.

2.Inversely proportional to the area of cross-section a. i.e. resistance decrease with increase in Area.

 

These above factors can be written mathematically as;

$R\infty L\text{       }\cdots \text{     (1)}$        

$R\infty \frac{1}{A}\text{       }\cdots \text{     (2)}$       

Comparing (1), (2) we can write

$R\infty \frac{L}{A}$

Or

$R=\rho \frac{L}{A}\text{     }\cdots \text{     (3)}$                   

Where  is the resistivity of the wire. It depends upon the nature of the material of a conductor. If resistivity is large, it is poor conductor and if resistivity is small, it is good conductor.

From (3)

$\rho =R\frac{A}{L}$

If L= 1m and A=1mthen,

$\rho =R$

So,

The resistivity of a substance is defined as the resistance of a meter cube of the substance. The unit of resistivity is ohm-meter.

Conductance

Sometimes, it is more convenient to express the ease with which current flows through the circuit rather than the amount of resistance encountered. The ease with which current flows is referred to as conductance, designated by G. it is the reciprocal of resistance. Thus, we see that any device or circuit has conductance as well as resistance. The unit of conductance is the Siemens (S).

Since conductance is the reciprocal of resistance, it is expressed mathematically as follows:

$G(Siemens)=\frac{1}{R(ohm)}$

Example

What is the conductance of a transistor with an internal resistance of 5000 Ω?

Solution

$G=\frac{1}{5000}=\frac{1}{5*{{10}^{3}}}$

$G=0.0002\text{ S=0}\text{.2 mS}$

Resistors in Series

The following figure shows three resistors are+ connected in series

Resistors in Series

Using Kirchhoff’s voltage law, a procedure can be developed to determine the total resistance. Applying Kirchhoff’s law to above figure, we obtain

${{V}_{T}}={{V}_{1}}+{{V}_{2}}+{{V}_{3}}\text{      }\cdots \text{      (4)}$

By definition, the same current has to flow through all elements connected in series. Thus iT flows through each resistance. Dividing both sides of equation (4) by iT we obtain

$\frac{{{V}_{T}}}{{{i}_{T}}}=\frac{{{V}_{1}}}{{{i}_{T}}}+\frac{{{V}_{2}}}{{{i}_{T}}}+\frac{{{V}_{3}}}{{{i}_{T}}}~~~~~~~~~~~~\cdots ~~~~~~~~~~~\left( 5 \right)$

Analyzing equation (5), we find that left-hand side yields the total resistance RT of the circuit using ohm’s law; equation (5) yields

${{R}_{T}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}~~~~~~\cdots ~~~~~~\left( 6 \right)$

Eq. (6) states that when resistors are connected in series, the total resistance is the sum of the individual resistances. If number of equal resistors are connected in series, we may use

${{R}_{T}}=NR$

Where N is the number of equal resistors and R is the value of one of the resistors.

Resistors in Parallel

Figure below shows three resistors connected in parallel

Resistors in Parallel

Applying Kirchhoff’s current law, we obtain

${{i}_{T}}={{i}_{1}}+{{i}_{2}}+{{i}_{3}}~~~~~~~\cdots ~~~~~~\left( 7 \right)$

Since VT is the voltage across each of the parallel resistors, we may divide eq. (7) by VT:

$\frac{{{i}_{T}}}{{{V}_{T}}}=\frac{{{i}_{1}}}{{{V}_{T}}}+\frac{{{i}_{2}}}{{{V}_{T}}}+\frac{{{i}_{3}}}{{{V}_{T}}}~~~~~~~~~~\ldots ~~~~~~~~\left( 8 \right)$

Analyzing both sides of eq. (8), we see terms are current divided by voltage. The left hand side yields the reciprocal of total resistance. The terms on right hand side yield;

$\frac{{{i}_{1}}}{{{V}_{T}}}=\frac{1}{{{R}_{1}}}$

$\frac{{{i}_{2}}}{{{V}_{T}}}=\frac{1}{{{R}_{2}}}$

$\frac{{{i}_{3}}}{{{V}_{T}}}=\frac{1}{{{R}_{3}}}$

Thus;

$\frac{1}{{{R}_{T}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}~~~~~\cdots ~~~~~~\left( 9 \right)$

We note that resistors in parallel combine in the same manner as capacitors in series. For any two resistors in parallel, we may use the product over the sum rule to determine the total resistance:

${{R}_{T}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$

If two or more parallel resistors are equal, then total resistance can be obtained as;

${{R}_{T}}=\frac{R}{N}$

Where N is the number of equal resistors and R is the value of one of parallel resistors.

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About Ahmad Faizan

Mr. Ahmed Faizan Sheikh, M.Sc. (USA), Research Fellow (USA), a member of IEEE & CIGRE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from Kansas State University, USA.