A) \[50\,\,\Omega ,\,4.4\,A\]
B) \[40.4\,\,\Omega ,\,5\,A\]
C) \[3.07\,\,\Omega ,\,6.0\,A\]
D) \[11.4\,\,\Omega ,\,17.5\,A\]
Correct Answer: A
Solution :
In an L-R circuit, \[{{V}_{R}}\]and \[{{V}_{L}}\], are mutually perpendicular to each other, hence impedance \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] \[\therefore \] \[Z=\sqrt{{{R}^{2}}+{{(\omega \,L)}^{2}}}\] Given, \[R=30\,\Omega ,\,\omega =2\pi n=2\pi \times 50\], \[L=\frac{0.4}{\pi }\] \[\therefore \] \[Z=\sqrt{{{(30)}^{2}}+{{\left( \frac{2\pi \times 50\times 0.4}{\pi } \right)}^{2}}}=50\,\Omega \] Current in the circuit is given by \[i=\frac{V}{2}=\frac{220}{50}=4.4\,A\]You need to login to perform this action.
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