In a series LCR circuit the frequency of a 10V- AC voltage source is adjusted in such a fashion that the reactanace of the inductor measures 15 - and that of the capacitor 11- If R - 3 - the potential difference across the series combination of L and C will be-

I=V/Z=V/√(R^2+[X_L X_C]^2)=10/√(3^2+[15 11]^2) =10/5=2A V_L V_C=I[X_L X_C]=2[15 11] = 8V ...

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Byju's Answer

Standard XII

Physics

RLC Circuit

In a series L...

Question

In a series LCR circuit the frequency of a 10V, AC voltage source is adjusted in such a fashion that the reactanace of the inductor measures 15 $Ω$ and that of the capacitor 11 $Ω$ ,. If R = 3 $Ω$ the potential difference across the series combination of L and C will be:

8 V

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10 V

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22 V

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52 V

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Solution

The correct option is A 8 V
$I = \frac{V}{Z} = \frac{V}{\sqrt{R^{2} + [X_{L} - X_{C}]^{2}}} = \frac{10}{\sqrt{3^{2} + [15 - 11]^{2}}}$
$= \frac{10}{5} = 2 A$
$V_{L} - V_{C} = I [X_{L} - X_{C}] = 2 [15 - 11]$
= 8V

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In a series LCR circuit the frequency of a 10V, AC voltage source is adjusted in such a fashion that the reactanace of the inductor measures 15 Ω and that of the capacitor 11Ω,. If R = 3 Ω the potential difference across the series combination of L and C will be:

The correct option is A 8 VI=VZ=V√R2+[XL−XC]2=10√32+[15−11]2 =105=2A VL−VC=I[XL−XC]=2[15−11] = 8V

In a series LCR circuit the frequency of a 10V, AC voltage source is adjusted in such a fashion that the reactanace of the inductor measures 15 $Ω$ and that of the capacitor 11 $Ω$ ,. If R = 3 $Ω$ the potential difference across the series combination of L and C will be:

The correct option is A 8 V
$I = \frac{V}{Z} = \frac{V}{\sqrt{R^{2} + [X_{L} - X_{C}]^{2}}} = \frac{10}{\sqrt{3^{2} + [15 - 11]^{2}}}$
$= \frac{10}{5} = 2 A$
$V_{L} - V_{C} = I [X_{L} - X_{C}] = 2 [15 - 11]$
= 8V