in the arrangement shown, readings of ideal voltmeters V1, V2 and V3 are 20 volt, 50 volt and 60 volt respectively. then current flowing through the current carrying branch is

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5 A

2.5 A

1 A

0.5 A

answer is C.

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Detailed Solution

V1=iP, V2=iQ+R and V3=iP+Q∴iR=V1+V2−V3⇒i=V1+V2−V3R=20+50−6010Amp ⇒i= 1 A

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