Combination of Resistors , Equivalent Resistance

Combination of Resistors :

(i) Series combination

(ii) Parallel combination

 Series combination

When resistances are connected in series,

(a) same current passes through each resistor.

(b)P otential differences are in the ratio V1 : V2 : V3 ……. = R1 : R2 : R3…..

$\large V = V_1 + V_2 + V_3 + …..$

$\large I R_{eq} = I R_1 + I R_2 + I R_3 + …$

$\large R_{eq} = R_1 + R_2 + R_3 + …. $

Note: When two resistances are connected in series then

$\large V_1 = \frac{V R_1}{R_1 + R_2} \; and \; V_2 = \frac{V R_2}{R_1 + R_2}$

 Parallel combination

When resistors are joined in parallel

(a) the potential difference across each resistor is same.

(b) the current are in the ratio  $\large I_1 : I_2 : I_3 …. = \frac{1}{R_1} : \frac{1}{R_2} : \frac{1}{R_3} …$

Since , I = I1 + I2 + I3 …. In

$ \displaystyle \frac{V}{R_{eq}} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} + …$

$ \displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + …$

Note: When two resistances are connected in Parallel then

$\large I_1 = \frac{I R_2}{R_1 + R_2} \; and \; I_2 = \frac{I R_1}{R_1 + R_2}$

Example : Find the equivalent resistance between A and B in the circuit shown here. Every resistance shown here is of 2 Ω.

Solution:  Due to Symmetry , Points C , O & D are at the same potential.

Therefore, resistances AO, AC and AD are in parallel .

$ \displaystyle \frac{1}{R_{AO}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$

$ \displaystyle \frac{1}{R_{AO}} = \frac{3}{2} $

RAO = 2/3 Ω

Similarly BC, BO and BD are in parallel.

RBO = 2/3 Ω

As, RAO  and RBO are in series ,

Hence , RAB = RAO  + RBO= 2/3 + 2/3

RAB = 4/3 = 1.33 Ω

Example : Six equal resistances each of resistance 4 Ω are connected to form the following figure. What is the resistance between any two corners.

Solution:

There is a symmetry about the line passing through QO and mid point of PR.

Numerical

$ \displaystyle \frac{1}{R_{PQ}} = \frac{1}{r} +\frac{1}{r} + \frac{1}{r/2} $

$ \displaystyle \frac{1}{R_{PQ}} = \frac{4}{r}$

RPQ = r/4

Similarly , RQR = r/4

As RPQ & RQR are in series

$ \displaystyle R_{eq} = \frac{r}{4} + \frac{r}{4}= r/2 $

Req = r/2 = 4/2 = 2 ohm

Example  : In the network shown in figure, each resistance is 1 Ω. What is the effective resistance between A and B.

Solution: There is a symmetry about line passing through E and mid point of CD.

RAB = 8r/7

Example : Find the equivalent resistance between points A & B of the network shown in the given diagram

Solution:

The resistors 3 Ω and 6 Ω are in series and so are 5 Ω and 10 Ω resistors. These two series equivalents are in parallel to each other and also to the 4 resistors. Hence the network reduces to the one given below :

Numerical

Req = 1 + 2.34 + 2

Req = 5.34 Ω

Example : An infinite ladder network of resistance is constructed with 1 Ω and 2Ω resistances, as shown in figure.
The 6V battery between A and B has negligible internal resistance.
(i) Show that the effective resistance between A and B is 2 Ω.
(ii) What is the current that passes through the 2 Ω resistance nearest to the battery ?

Infinite Ladder

Solution:
Let the effective resistance between points C and D be R, then the circuit can be redrawn as shown in figure.

Infinite Ladder

The effective resistance between A and B is

$ \displaystyle R_{eq} = 1 + \frac{2 \times R}{2+R}$

This resistance Req can be taken as R because if we add one identical item in infinite item, then the result will be the same. Therefore,

$ \displaystyle 1 + \frac{2 \times R}{2+R} = R $

R2 – R – 2 = 0

(R+1)(R-2) = 0

Hence , R = 2 Ω

Current i = 6/2 = 3 A

Current passes through the 2 Ω resistance = i/2 = 1.5 A

Example : Compute the equivalent resistance of the network shown in figure and find the current i drawn from the battery.

Solution: The 6Ω and 3Ω resistances are in parallel. Their equivalent resistance is,

$\large \frac{1}{R’} = \frac{1}{6} + \frac{1}{3}$

R’ = 2 Ω

Now, this 2Ω and 4Ω resistances are in series and their equivalent resistance is R = 4 + 2 = 6Ω .

Current drawn from the battery is $\large i = \frac{Total \; emf}{Total \;resistance}$

$\large i = \frac{18}{6} = 3 A$

Also Read :

→ Electric Current , Drift velocity & Deduction of Ohm’s Law
→ Grouping of Cells
→Concept of EMF & Terminal P.D
→Kirchhoff ‘ s Laws & Solved Examples
→Wheatstone Bridge Principle & Metre Bridge
→Ammeter and Voltmeter
→Potentiometer & its Applications
→Sensitivity of Potentiometer
→Charging of capacitors
→Discharging of Capacitors
→ Heating effect of currents

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