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Four resistances of 15\Omega ,12\Omega,4\Omega\:\:and\:\:10\Omega respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10\Omega to balance the network is ___________\Omega.
Option: 1 10
Option: 2 5
Option: 3 15
Option: 4 20

Answers (1)


For balanced Wheatstone bridge \frac{R_1}{R_2}=\frac{R_3}{R_4} \ \ or \ \ \frac{R_1}{R_3}=\frac{R_2}{R_4}

R_2=12 \Omega \ \ and \ \ R_4=4 \Omega

As \frac{R_2}{R_4}=\frac{12}{4}=3

So using R_1=15 \Omega

We get R_3=R_{AD}=5 \Omega 

let we connected x-ohms in parallel to 10-ohm resistance

i.e R_3=5 \Omega=\frac{x*10}{x+10}

we get x=10 \Omega 


So the answer will be 10.

Posted by

vishal kumar

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