Equivalent resistance [![enter image description here][2]][2]
Can any one please explain how in the solution of question above they have made the first transformation of circuit by removing the 8 ohm resistor keeping the whole circuit same. How can 8 ohm resistor be removed directly and is there any trick to solve such type of problems.
 A: If you look at the 5 resistors in that initial group you will notice that the ratio of the resistors in each of the legs without the removed one is the same.  (1:2)  Now consider the voltage from one end to the other.  The voltage drop across the first resister will be the same in both legs.  Likewise the voltage drop across the second resistors will be the same.  What this means is that the voltage across the removed resistor will be 0.  Therefore there is no current going through that resistor and it can be removed with no change to the functional circuit.
A: The sub-network ABCD is a Wheatstone Bridge arrangement. The condition for 'balance' - ie no current through BD - is that $R_{AB}/R_{BC} = R_{AD}/R_{DC}$. Conversely, if the resistors in the circuit are in this ratio, then there is no current in BD. The value of $R_{BD}$ does not make any difference to the currents through ABC and ADC, so it can be removed from the network.

The condition can be derived by noting that, if no current flows through BD, then B and D must be at the same voltage. Since the PD along ABC is the same as that along ADC, this means that we must have $V_{AB}/V_{BC}=V_{AD}/V_{DC}$. Since the currents in AB,BC are the same, and the currents in AD,DC are the same, and $V=IR$, this ratio is equivalent to $R_{AB}/R_{BC} = R_{AD}/R_{DC}$.
