Question

solve this question ​

Answer 1

Answer:

The resistance between A and B of the given network is 2/3 Ω

Explanation:

Lets divide the given circuit into 3 parts as shown in the figure below:

In part 1, we have two resistor of 1 Ω in series

Hence for measuring series resistance we have,

Rs = r1 + r2 + r3 + ...... + rn

where,

n is the number of resistor

Rs is the resultant resistance in series

here for part 1

Rs= 1+1 = 2Ω

Now, between point A and B, the resistors in Part A, B and C are in parallel

and the value of each resistor in parallel is 2Ω each in each parts

For parallel resistance we have,

1/Rp = 1/r1 + 1/r2 + 1/r3 + ..... + 1/rn

Where,

n is the number of resistors

Rp is the resultant resistance in parallel

here we have,

1/Rp= 1/2 + 1/2 + 1/2

1/Rp = 3/2

Rp= 2/3 Ω