The value of ′Z′ such that no current flows in branch Y will be
A
(3.46+j2)Ω
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B
(13.85+j8)Ω
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C
(6.93+j4)Ω
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D
(27.7+j16)Ω
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Solution
The correct option is C(6.93+j4)Ω
Let, VRY=V∠0∘V VYB=V∠−120∘V VBR=V∠120∘
Let current through capacitor be IC IC=V∠−120∘Xc=V∠−120∘−j8 ∴IC=V8∠−120∘+90∘ IC=V8∠−30∘A
Let current though phase Y be IY ∴IY=IC−IL
as IY=0 IC=IL ∴V8∠−30∘=V∠0∘R+jX∴R+jX=8∠30∘=8∠30∘ΩZ=(6.93+j4)Ω