Question

find the effective resistance between the point A and B in the network shown in the figure.

Answer 1

answer will be 4_Π_ ......

Answer 2

Answer:

The net effective resistance is 4Ω.

Solution:

Let, the three resistors be named as R1, R2 and R3. So ‘R2 and R3’ are connected in ‘parallel’ and ‘R1 is connected in series’ with R2 and R3. So first we have to find the ‘effective resistance’ of resistors connected in parallel and then the net effective resistance in series.

Let R1 = 2 ohms, R2 = 6 ohms and R3 = 3 ohms

The effective resistance in parallel connection is  

$\frac{1}{R_{p}}=\frac{1}{R 2}+\frac{1}{R 3}=\frac{1}{6}+\frac{1}{3}$

$\frac{1}{R_{p}}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}$

$\bold{\therefore R_{p}=2}$

Then the net effective resistance can be found by finding the ‘effective resistance’ in series connection between $\mathrm{R} 1 \text { and } \mathrm{R}_{\mathrm{p}}$.

$R_{s}=R 1+R_{p}=2+2=4$

Thus, the net ‘effective resistance’ is 4Ω.