Electricity Worksheet-7

Electricity Worksheet-7


  1. Calculate the equivalent resistance of the network across the points A and B shown in the figure.


  1. Three resistors are connected as shown in the figure. Through the resistor of 5 Ω, a current of 1 A is flowing.

(a) Find the potential difference across AB and across AC?

(b) Find the current through the other two resistors?

(c) Find the total resistance?


  1. A 24 volt battery is connected to the arrangement of resistances shown in fig.  calculate (i) the total effective resistance of the circuit, (ii) the total current flowing in the circuit.


  1. Three resistors of 6 Ω, 3 Ω and 2 Ω are connected together so that the total resistance is greater than 6 Ω but less than 8 Ω. Draw a diagram to show this arrangement and calculate its total resistance.


  1. For the combination of resistors shown in the figure, find the equivalent resistance between (i) C and D and (ii) A and B.




The given network, shown in the figure a, can be represented as shown in figure b.

The resultant of 4 Ω and 8 Ω resistances, which are in series in the arm EDC, is given by

          R1 = 4 Ω + 8 Ω = 12 Ω

Similarly, the resultant of 4 Ω and 8 Ω in the arm EFC is given by

          R2 = 4 Ω + 8 Ω = 12 Ω

The equivalent circuit of R1 and R2 which are in parallel is shown in figure c.                                                     

If R is the equivalent resistance between the point A and B, then

Or     R = 6 Ω


  1. (a)    As current (I2) through R1 is 1 A,

          Potential difference across AB = V1 = I1R1 = 1 × 5 = 5V

          Since R2 and R3 are in parallel, resultant resistance between B and C is given by


          Potential difference across BC, i.e.,

V2 = I1 RP = 1 × 6 = 6V (∵ current through BC is I1, i.e., 1 A)

          Potential difference  across AC

         = Potential difference across AB + Potential difference across BC = 5 V + 6 V = 11 V

(b) current through R2, i.e.,

       (∵ Potential difference  across R2 is the same as across BC)       

          Current through , i.e., I3 = I1 – I2 = 1A – 0.6 = 0.4 A

 (c) Total resistance between A and C, i.e., R = R1 + RP = 5Ω + 6Ω = 11Ω

          The resultant of R2 and R3 which are in parallel is RP; and R1 and RP are in series


  1. R1 is the equivalent resistance of 5Ω and 20Ω which are in series,

          R1 = 5Ω + 20Ω = 25Ω


          R2 is the equivalent resistance of 15Ω and 10Ω which are in series,

          R2 = 15 Ω + 10 Ω = 25 Ω

          Now R1 and R2 are in parallel so final resistance is given by


Total potential difference across the arrangement of all the resistances is 24 V

          Total current in the circuit,


  1. Here the total resistance has to be greater than 6Ω, therefore, 6Ω resistance should be in series with some other resistance. Now the total resistance is less than 8Ω, the other two resistances should be in parallel and their combination be connected in series with 6Ω resistance. Now RP is the resultant of two resistances of 3Ω and 2Ω in parallel, then



Since RP and 6Ω are in series, total resistance of the combination, i.e.,

          R = 6 + RP = 6 + 1.2 = 7.2Ω


  1. (i)     The resistors R2, R3 and R4 are in series. Their equivalent resistance Rs is given by

          Rs = 3Ω + 3Ω + 3Ω  = 9Ω

          Further, Rs (= 9Ω) and R5 (= 3Ω) are in parallel so their equivalent resistance is given by




          Thus, the equivalent resistance between C and D is 2.25 Ω.

(ii)    Now R1 (= 3Ω), RP (= 2.25 Ω)

          and   R6 (= 3 Ω)

          Are in series

Thus, equivalent resistance between A and B,

          = R1 + RP + R6

          = 3 + RP + R6

          = 3 + 2.25 + 3

          = 8.25 Ω