A silicon diode has a "knee voltage of" threshold voltage" value of 0.7V. This Si diode is connected in a circuit as shown in the figure
Potential $V_{0}$ of point p in the circuit will be:

-0.44V

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0.44V

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4.4V

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-4.4V

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Solution

The correct option is A -0.44V
By KCL-
$\frac{10 - (V_{0} + 0.7)}{4.7} + \frac{- 5 - V_{0}}{2.2} = - 0.44$

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