JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 17)
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As per the given graph, choose the correct representation for curve $$\mathrm{A}$$ and curve B.
Where $$\mathrm{X}_{\mathrm{C}}=$$ reactance of pure capacitive circuit connected with A.C. source
$$\mathrm{X}_{\mathrm{L}}=$$ reactance of pure inductive circuit connected with $$\mathrm{A} . \mathrm{C}$$. source
R = impedance of pure resistive circuit connected with A.C. source.
$$\mathrm{Z}=$$ Impedance of the LCR series circuit $$\}$$
$$\mathrm{A}=\mathrm{X}_{\mathrm{L}}, \mathrm{B}=\mathrm{R}$$
$$\mathrm{A}=\mathrm{X}_{L}, \mathrm{~B}=Z$$
$$\mathrm{A}=\mathrm{X}_{\mathrm{C}}, \mathrm{B}=\mathrm{X}_{\mathrm{L}}$$
$$\mathrm{A}=\mathrm{X}_{\mathrm{C}}, \mathrm{B}=\mathrm{R}$$
Explanation
$$
\begin{aligned}
& \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{(2 \pi \mathrm{f}) \mathrm{C}} \\\\
& \therefore \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}} \\\\
& \therefore \text { Curve } \mathrm{A} \\\\
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=(2 \pi \mathrm{f}) \mathrm{L} \\\\
& \therefore \mathrm{X}_{\mathrm{L}} \propto \mathrm{f} \\\\
& \therefore \text { Curve } \mathrm{B}
\end{aligned}
$$
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