MCQ
In a series $LCR$ current the voltage across resistance, capacitance and inductance is $10\, V$ each. If the inductor is short circuited, the voltage across the capacitor will be:-
- A$10\,V$
- BZero
- C$10 \sqrt 2 \,V$
- ✓$\frac{10}{\sqrt 2} \,V$
✓
Answer
Correct option: D.
$\frac{10}{\sqrt 2} \,V$
d
As $\mathrm{V}_{\mathrm{L}}=\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{R}}$ So $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}=\mathrm{R}$
As $\mathrm{V}_{\mathrm{L}}=\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{R}}$ So $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}=\mathrm{R}$
and $\mathrm{E}=\mathrm{V}_{\mathrm{R}}=10 \mathrm{\,V}$
Now $\mathrm{i}=\frac{\mathrm{E}}{\mathrm{Z}} \Rightarrow \mathrm{i}=\frac{10}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{C}}^{2}}}=\frac{10}{\sqrt{2} \mathrm{R}}$
$\mathrm{V}_{\mathrm{C}}=\mathrm{iX}_{\mathrm{C}}=\frac{10}{\sqrt{2} \mathrm{R}} \times \mathrm{R}=\frac{10}{\sqrt{2}} \mathrm{\,Volt}.$
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