In the circuit shown in figure, the potentials of B,C and D are :
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a
VB=6V;VC=9V;VD=11V
b
VB=11V;VC=9V;VD=6V
c
VB=9V;VC=11V;VD=6V
d
VB=9V;VC=6V;VD=11V
answer is B.
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Detailed Solution
Potential at O is zero being earthed.Applying Kirchhoff’s second lawi(1+ 2 + 3) =12 - 6 or i = 1AVA−VD=(1+2+3)×1=6VVA−VB=1×1=1VVA−VC=(1+2)×1=3VAlso, VA−VO=12V or VA=12VThus, VD=12−6=6VVB=12−1=11V,VC=12−3=9V