Question

In a typical Wheatstone network the resistance in cyclic order are $A=10\Omega ,B=5\Omega ,C=4\Omega$ and $D=4\Omega$ for the bridge to be balanced :-

• A. $10\Omega$ should be connected in parallel $A$
• B. $10\Omega$ should be connected in series with $A$
• C. $5\Omega$ should be connected in series with $B$
• D. $5\Omega$ should be connected in parallel with $B$

Answer 1

The correct option is C $5\Omega$ should be connected in series with $B$

a) here, $A=10||10=10/2=5,B=5,C=4,D=4$
and $A/B=1,C/D=1$ so, $A/B=C/D$
b) here, $A=10+10=20,B=5,C=4,D=4$
and $A/B=4,C/D=1$ so$,A/B\ne C/D$
c) here,$A=10,B=5+5=10,C=4,D=4$
and $A/B=1,C/D=1$ so, $A/B=C/D$
d)here, $A=10,B=5||5=5/2=2.5,C=4,D=4$
And $A/B=4,C/D=1$ so,$A/B\ne C/D$