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Resistances in Series and Resistances in Parallel

Resistances in Series

When some conductors having resistances R1, R2, and R3 etc. are joined end on end as in figure below they are said to be connected in series. It can be proved that the equivalent resistance or total resistance between points A and D is equal to the sum of the three individual resistances. Being a series circuit, it should be kept in mind that

figure 1


figure 2

(i) current is the same through all three conductors  (I = I1 = I2 = I3)
(ii) voltage drop across each is different due to Its resistance and is given by Ohm's Law
(iii) sum of the three voltage drops is equal to the voltage applied across the three conductors. 

There is a progressive fall in potential as we go from point A to D as shown in Figure 3
figure 3


 V = V1 + V2 + V3 = IR1 + IR2 + IR3

But V = IR

where R is the equivalent resistance of the series combination.

IR=IR1 + IR2 + IR3

Req=R1+R2+R3


Resistances in Parallel

Three resistances, as joined in Figure 4, are said to be connected in parallel. In this case
(i) potential differen (V) across all resistances is the same
(ii) current in each resistor is different and is given by Ohm's Law and
(iii) the total current is the sum of the three separate currents

 
figure 4


I = I 1  + I2 + I3 = V/R1 + V/R2 + V/R3

I = V/R 

where V is applied voltage

R = equivalent resistance of the parallel combination.

 

V/R = V/R1 + V/R2 + V/R3 

OR

1/R = 1/R1 + 1/R2 + 1/R3

 Example 1

Calculate the effective resistance of the following combination of resistances and the voltage drop across each resistance when a P.D (potential different) of 60 V is applied between points A and B.

figure 5

Solution:

RESISTANCE

    (i) Resistance betweet A dan C

        in this circuit we use parallel resistance formula

        RAC = (3 x 6)/(3+6) = 18/9 = 2 Ω

    (ii) Resistance of branch ACD

        in this circuit we use series resistance formula

        RACD =  RAC +  RCD =  2 + 10  = 12 Ω

    (iii) Resistance betweet A dan D

        in this circuit we use parallel resistance formula

        now there are two parallel path between point A and D of resistance 12 Ω and 6 Ω. hence,

        resistancce point A and D 

        RAD =  (12 x 6)/(12+6) = 4 Ω 

***** Resistance between A dan B *** 

   in this circuit we use parallel resistance formula

        RAB = RAD +  RDB = 4 + 6 = 10 Ω  


CURRENT 

    (i) total circuit current 

        Itotal = V/RAB = 60 / 10 = 6 A

    (ii) current through 6 Ω (R5 ­) resistor  

        IR5  = Itotal  x RACD  / (RACD +  R5)

            = 6 x 12/18 

            = 4 A

 (iii) current in branch ACD

        IACD  = Itotal  x R5 ­/ (RACD +  R5)

                 = 6 x 6/18

                = 2 A

POTENTIAL DIFFERENT 

(i) PD across R1  and  R2 = IACD  x RAC = 2 x 2 = 4 V

(ii) PD across R3  = IACD  x RCD = 2 x 10 = 20 V

(iii) PD across R4  =  Itotal x RDB = 6 x 6 = 36 V

(iv) PD across R5  =  IR5  x R5  = 4 x 6 = 24 V


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