Question

For the network shown in the figure find the value of current i ?

options:-
9v/35
18v/5
5v/9
5v/18

Answer 1

Here the ratio of resistances at the sides of the square net work suggest that the connection follows Wheatstone bridge network. So the removal of 4ohm resistor at the diaginal will not make any change of equivalent resistance of the network where 4ohm and 2ohm in serises together will form 6ohm and that combination is connected in parallel with 9ohm formed by series combination of 6ohm and 3 ohm resistors.

Hence equivalent resistance become

`R=1/(1/6+1/9)=18/5Omega`

So the main current in the circuit

`I="emf"/"total resistance "=V/(18/5)=(5V)/18A`