For the circuit shown below,
$V_{0}$ is

12 V

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4 V

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48 V

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6 V

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Solution

The correct option is C 48 V
We apply KVL in the loop,
$- 12 + 4 i + 2 V_{0} - 4 + 6 i = 0$
Put, $V_{0} = - 6 i$
$- 12 + 4 i - 12 i - 4 + 6 i = 0$
$- 2 i = 16$
$i = - 8 A$
$V_{0} = - 6 \times - 8 = 48 V$

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