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In $R-L-C$ series circuit, the potential differences across each element is $20 \mathrm{~V}$. Now the value of the resistance alone is doubled, then P.D. across $R, L$ and $C$ respectively.
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The correct answer is:
$20 \mathrm{~V}, 10 \mathrm{~V}, 10 \mathrm{~V}$
Given, the potential difference across each element is same i.e., $20 \mathrm{~V}$, so, the circuit is at resonance and we have
$V=V_{R}=20 \mathrm{~V}$
If the value of $R$ is doubled, then value of / reduced to half $\therefore$
$V_{L}=V_{C}=\frac{I X_{L}}{2}=\frac{I X_{C}}{2}=\frac{20}{2}=10 \mathrm{~V}$
$(\because$ circuit is at resonance) and $\quad V_{R}=V=20 \mathrm{~V}$
$V=V_{R}=20 \mathrm{~V}$
If the value of $R$ is doubled, then value of / reduced to half $\therefore$
$V_{L}=V_{C}=\frac{I X_{L}}{2}=\frac{I X_{C}}{2}=\frac{20}{2}=10 \mathrm{~V}$
$(\because$ circuit is at resonance) and $\quad V_{R}=V=20 \mathrm{~V}$
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