Find the equivalent resistance of the network shown in figure between the points `a and b`.

Explanation

Suppose a current I enters the network at point a and the same current leaves it at point b. Supposem the currents in ac, ad and ae are`i_(1),i_(2)and i_(3)` respectively. Similar will be the distribution of current at b. The current I leaving at b is composed if `i_(1)`form db, `i_(2)`from cb and`i_(3)`from eb. The situation is shown in figure.
As the current in ae is equal to that is eb, the current in ce will equal to the current in ed fron the junction law. If we assume that the branches ced ad aeb do not physically touch at e,nothing will be changed in the current distribution.We can then represent the branch aeb by a single resistance of 10(Omega) connected between a and b. Similarly, the branch ced may be replaced a single `5(Omega)`resistor between c and d. The circuit is redrawn in figure.This is same as the circuit in figure conected in parallel with a resistance of `10(Omega)`.So the network is equivalent to a parallel conbination of `7(Omega)and 10(Omega)` resistor. The equivalent resistance of the whole network is, therefore,
`R=((7omega)xx(10Omega))/(7Omega+10Omega)=4.1(Omega).`