Find the root mean square value of voltage $V_{X}$ in the figure shown.

15.53 V

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7.83 V

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11.83 V

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2.88 V

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Solution

The correct option is C $11.83 V$
Step 1: Deactivate $20 V$ source by short circuit

$\frac{V_{X}}{1 / j w C} + \frac{V_{X}}{2} + \frac{V_{X} - 20}{2} = 0$

Given that, $ω = 5 r a d / s e c$ and $C = 0.1 F$

$j 0.5 V_{X} + \frac{V_{X}}{2} + \frac{V_{X}}{2} = 10$

$V_{X} [1 + j 0.5] = 10$

$| V_{X} | = \frac{10}{1 + j 0.5} = \frac{10}{\sqrt{1^{2} + {0.5}^{2}} ∠ {26.56}^{o}} = 8.94 ∠ - {26.56}^{o}$

$V_{X} = 8.94 s i n (5 t - {26.56}^{o})$ V

Step 2: Deactivate $20 s i n 5 t$ source by short circuit

When $D C$ is connected capacitor becomes open circuit

$V_{X} = 20 \times \frac{2}{4} = 10 V$

$∴$ Net voltage,

$V_{X} = 10 + 8.94 s i n (5 t - {26.56}^{o})$

$V_{X (r m s)} = {⎛ ⎝ 10^{2} + {(\frac{8.94}{\sqrt{2}})}^{2} ⎞ ⎠}^{1 / 2}$

$= 11.83 V$

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