ICSE Class 10 Answered

Fig.1 gives the given network of resistrs .

We see that in the loop AFEA , series combination of two 2Ω resistors is in parallel to 4Ω resistor .

Equivalenet resistance of series combination of two 2Ω resistors is 4Ω .

Hence if  parallel series combination of two 2Ω resistors is in parallel to 4Ω resistor then net equivalenet resistance is 2Ω.

Hence , in fig.2 , we have only one 2Ω resistance across the points A and E.

Again in fig.2, we see that in the loop AEDA series combination of two 2Ω resistors is in parallel to 4Ω resistor .

As explained before , we get net equivalenet resistance is 2Ω .

Hence , in fig.3 , we have only one 2Ω resistance across the points A and D.

Again in fig.3, we see that in the loop ADCA series combination of two 2Ω resistors is in parallel to 4Ω resistor .

As explained before , we get net equivalenet resistance is 2Ω .

Hence , in fig.4 , we have only one 2Ω resistance across the points A and C.

Finally  in fig.4, we see that in the loop ACBA series combination of two 2Ω resistors is in parallel to 4Ω resistor .

As explained before , we get net equivalenet resistance is 2Ω .

Hence , equivalenet resistance across A and B is 2Ω .