With sources $i_{A} a n d V_{B}$ are on in the circuit given below then $V_{C} = 0, i_{x} = 20 A$ ,and when sources $i_{A} a n d V_{C}$ are on then $V_{B} = 0, i_{x} = - 5$ and finally, with all three sources on, $i_{x} = 12 A .$ If the the only source $V_{B}$ is operating, then the value of $i_{x}$ is

17 A

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-8 A

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32 A

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7 A

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Solution

The correct option is A 17 A
Let, $i_{x} = i_{x A} + i_{x B} + i_{x C}$
$i_{x A} + i_{x B} = 20$
$i_{x A} + i_{x c} = - 5$
$i_{x A} + i_{x B} + i_{x C} = 12$
$i_{x A} = 3 A;$
$i_{x B} = 17 A;$
$i_{x C} = - 8 A$
$∴$ if only source $V_{B}$ is operating,
then $i_{x} = i_{x B} = 17 A$

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With sources iAandVB are on in the circuit given below thenVC=0,ix=20A,and when sources iAandVC are on then VB=0,ix=−5 and finally, with all three sources on, ix=12A. If the the only source VB is operating, then the value of ix is

The correct option is A 17 ALet, ix=ixA+ixB+ixC ixA+ixB=20 ixA+ixc=−5 ixA+ixB+ixC=12 ixA=3A; ixB=17A; ixC=−8A ∴ if only source VB is operating, then ix=ixB=17A

The correct option is A 17 A
Let, $i_{x} = i_{x A} + i_{x B} + i_{x C}$
$i_{x A} + i_{x B} = 20$
$i_{x A} + i_{x c} = - 5$
$i_{x A} + i_{x B} + i_{x C} = 12$
$i_{x A} = 3 A;$
$i_{x B} = 17 A;$
$i_{x C} = - 8 A$
$∴$ if only source $V_{B}$ is operating,
then $i_{x} = i_{x B} = 17 A$