A series $L C R$ circuit is connected to a source of alternating emf $50 V$ and if the potential differences across inductor and capacitor are $90 V$ and $60 V$ respectively, the potential difference across resistor is:

400 V

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40 V

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80 V

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1600 V

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Solution

The correct option is A $40 V$
The potential difference across inductor and capacitor will be opposite to each other and that of the resistor will be perpendicular to them. The resultant of three must be $50 V$ .
As we know, $V = \sqrt{{(V_{L} - V_{C})}^{2} + {V_{R}}^{2}}$
Hence,
$\sqrt{{(90 - 60)}^{2} + V_{R}^{2}} = 50$

$900 + V_{R}^{2} = 2500$

$V_{R}^{2} = 1600$

$V_{R} = 40 V$

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A series LCR circuit is connected to a source of alternating emf 50 V and if the potential differences across inductor and capacitor are 90 V and 60 V respectively, the potential difference across resistor is:

The correct option is A 40 VThe potential difference across inductor and capacitor will be opposite to each other and that of the resistor will be perpendicular to them. The resultant of three must be 50V.As we know, V=√(VL−VC)2+VR2 Hence,√(90−60)2+V2R=50900+V2R=2500V2R=1600VR=40 V

A series $L C R$ circuit is connected to a source of alternating emf $50 V$ and if the potential differences across inductor and capacitor are $90 V$ and $60 V$ respectively, the potential difference across resistor is: