Consider the circuit shown in the figure below:
The voltage 'V' across the $1 Ω$ resistor is

- 5 V

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- 20 V

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- 6.67 V

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10 V

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Solution

The correct option is A $- 5 V$

Applying KCL at node (a) , we get ,

$\frac{V_{a}}{3} + \frac{V_{a}}{3} = 5 - 15$

$6 V_{a} = - 90$ $\Rightarrow V_{a} = - 15 V$

$∵ The current through 1 Ω$ resistor is $\frac{V_{a}}{3} = - \frac{15}{3} = - 5 A$

$∴ The voltage across 1 Ω resistor is - 5 V$

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