Grouping Of Resistances
Current Electricity of Class 12
Series Combinations
Let the equivalent resistance between A and B equals Req , By definition, Req =V/I. . . (1) 

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
Parallel Combinations Here again, Req =V/I . . .(1) I = i1 + i2 + i3 = . . . (2) From (1) and (2)



Illustration 22. Calculate the equivalent resistance between points A and E as shown in the figure. Each resistance is of 2 Ω. 

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.
Illustration 23. A battery of emf 10 V is connected to resistances as shown in the figure. Determine the potential difference between A and B. 

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.
Illustration 24. Find the effective resistance between the points A and B. 

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
For a certain adjustment of Q, VBD = 0, then no current flows through the galvanometer. ⇒ VB = VD or VAB= VAD ⇒ I1.P = I2.R Likewise, VBC = VDC ⇒ I1.Q = I2.S Dividing, we get, P/Q = R/S 

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

(b)

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
 Electric Current
 Mechanism Of Current Flow In Metallic Conductors
 Ohm's Law
 Specific Resistance Of The Material Of A Wire
 Measurement Of Unknown Resistance Using A Post Office Box
 Classification Of Materials In Terms Of Conductivity
 Kirchhoff Law
 Grouping Of Resistances
 Grouping Of Identical Cells
 RC−Circuit
 Measuring Instruments
 Verification Of Ohm’s Law Using Voltmeter And Ammeter
 Potentiometer
 Energy, Power And Heating Effect Of The Current
 solved question
 Exercise 1
 Exercise 2