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Byju's Answer

Standard XII

Physics

RLC Circuit

The r.m.s. va...

Question

The r.m.s. value of potential difference V shown in the figure is :

V_{0} / \sqrt{3}

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V_{0}

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V_{0} / \sqrt{2}

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V_{0} / 2

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Solution

The correct option is C $V_{0} / \sqrt{2}$
$V_{r m s} = {[\frac{1}{T} \int_{0}^{T / 2} V_{0}^{2} d t]}^{1 / 2}$
$= {[\frac{V_{0}^{2}}{T} [t]_{0}^{T / 2}]}^{1 / 2}$
$= {[\frac{V_{0}^{2}}{T} (T / 2)]}^{1 / 2}$
or $V_{r m s} = {[\frac{V_{0}^{2}}{2}]}^{1 / 2} = \frac{V_{0}}{\sqrt{2}}$

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The r.m.s. value of potential difference V shown in the figure is :

The correct option is C V0/√2Vrms=[1T∫T/20V20dt]1/2=[V20T[t]T/20]1/2=[V20T(T/2)]1/2 or Vrms=[V202]1/2=V0√2

The correct option is C $V_{0} / \sqrt{2}$
$V_{r m s} = {[\frac{1}{T} \int_{0}^{T / 2} V_{0}^{2} d t]}^{1 / 2}$
$= {[\frac{V_{0}^{2}}{T} [t]_{0}^{T / 2}]}^{1 / 2}$
$= {[\frac{V_{0}^{2}}{T} (T / 2)]}^{1 / 2}$
or $V_{r m s} = {[\frac{V_{0}^{2}}{2}]}^{1 / 2} = \frac{V_{0}}{\sqrt{2}}$