Grouping Of Resistances

Series Combinations

Using Kirchoff's 2^nd rule for the loop shown in figure,

V = IR1 + IR2 + IR3 . . . (2)

From (1) and (2),    Req = R1 + R2 + R3

Solution: The points B, C and D are at the same potential.  So, resistances AB, AC and AD are in parallel. Similarly, the resistances BE, CE and DE are in parallel.  So, an equivalent of the given network is as under.

Parallel combination of 2 Ω, 2 Ω and 2 Ω gives 2/3Ω.

∴  RAE = 2 × 2/3 Ω = 4/3 Ω

= 1.33 Ω.

Solution: Total resistance = 4 x 4/4 + 4 = 2Ω

Current I = 10V/2Ω = 5A

Since the resistances of both the branches are equal, therefore the current of 5 A shall be equally distributed. 

Current through each branch = 5/2 A = 2.5A

Vc – VA = 2.5 × 1 = 2.5 V

Vc – VB = 2.5 × 3 = 7.5 V.

VA – VB = (Vc – VB) – (Vc – VA) = 7.5 – 2.5 = 5.0 V.

Solution: Resistors AF and FE are in series with each other. Therefore, network AEF reduces to a parallel combination of two resistors of 6 Ω each.

R = 6 x 6/6+6 = 3Ω.

Similarly, the resistance between A and D is given, 6 x 6/6 + 6 = 3 Ω.

Now, resistor AC is in parallel with the series combination of AD and DC. Therefore, the resistance between A and C is 6 x 6/6 + 6= 3 Ω.

AC + CB = 3 + 3 = 6 Ω, since they are in series.  

Resistance between A and B is given by,  1/R = 1/6 + 1/3 = 3/6 or R_AB = 2Ω

Wheatstone Bridge

Illustration 25. What's the effective resistance of following circuits?

Solution: (a)It is a Wheatstone bridge that is balanced. Hence, the central resistance labeled 'C' can be assumed as ineffective.   

⇒   Req = R.

(b) The resistor R is in parallel with a balanced Wheatstone bridge. 

⇒  Req. = R.R/R + R = R/2

Related Topics

• Grouping of Identical Cells
• RC-Circuit