A) \[\sqrt{{{R}^{2}}+2{{\pi }^{2}}{{v}^{2}}{{L}^{2}}}\]
B) \[\sqrt{{{R}^{2}}+4{{\pi }^{2}}{{v}^{2}}{{L}^{2}}}\]
C) \[\sqrt{R+4{{\pi }^{2}}{{v}^{2}}{{L}^{2}}}\]
D) \[\sqrt{R+2{{\pi }^{2}}{{v}^{2}}{{L}^{2}}}\]
Correct Answer: B
Solution :
Key Idea: Impedance 15 the effective resistance of L-R series circuit. Let alternating emf be applied to a circuit containing L and R in series. \[\therefore \] \[{{E}^{2}}=V_{R}^{2}+V_{L}^{2}\] \[{{E}^{2}}={{i}^{2}}{{R}^{2}}+X_{L}^{2}\] \[\Rightarrow \] \[i=\frac{E}{\sqrt{{{R}^{2}}+X_{L}^{2}}}\] Applying Ohm's law we see that \[\sqrt{{{R}^{2}}+X_{L}^{2}}\]is the effective resistance of the circuit. It is known as impedance Z \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] Also, \[{{X}_{L}}=\omega L\] \[\therefore \] \[Z=\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}\] Here \[\omega =2\pi v\] \[\therefore \] \[Z=\sqrt{{{R}^{2}}+{{(2\pi vL)}^{2}}}\] \[Z=\sqrt{{{R}^{2}}+4{{\pi }^{2}}{{v}^{2}}{{L}^{2}}}\]You need to login to perform this action.
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