A) \[1500\text{ }\Omega \]
B) \[1700\text{ }\Omega \]
C) \[2102\text{ }\Omega \]
D) \[2500\text{ }\Omega \]
Correct Answer: C
Solution :
Impedance\[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] \[{{X}_{L}}=\omega L=2\pi fL\] Given, \[R=2100\,\Omega ,f=50\,Hz,L=\frac{1}{\pi }\] \[Z=\sqrt{{{(2100)}^{2}}+{{(2\times 50)}^{2}}}\] \[=\sqrt{{{(2100)}^{2}}+{{(100)}^{2}}}\] \[Z=2102\,\Omega \]
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