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In a series \(L C R\) circuit the frequency of a \(10 \mathrm{~V}\) a.c. voltage source is adjusted in such a fashion that the reactance of the inductor measures \(15 \Omega\) and that of the capacitor \(11 \Omega\). If \(\mathrm{R}=3 \Omega\), the potential difference across the series combination of \(L\) and \(C\) will be:
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Verified Answer
The correct answer is:
\(8 \mathrm{~V}\)
\(\mathrm{X}_{\mathrm{L}}=15 \Omega ; \mathrm{X}_{\mathrm{C}}=11 \Omega ; \mathrm{R}=3 \Omega\)
\(\begin{aligned}
& \text { Impedance, } \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2} \\
& =\sqrt{9+16}=5 \Omega
\end{aligned}\)
\(\begin{aligned}
& \text { Impedance, } \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2} \\
& =\sqrt{9+16}=5 \Omega
\end{aligned}\)
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