# Equivalent resistance

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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.

• This question is unclear - literally : the image is not in focus. – sammy gerbil Sep 21 '16 at 21:55
• Ok I will replace this image – Avi Sep 23 '16 at 7:03

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 $R_{AB}/R_{BC} = R_{AD}/R_{DC}$.