question_answer

In an L-R circuit \[L=\frac{0.4}{\pi }\] and R = 30\[\Omega \] If the circuit has an alternating emf of 220 V at 50 cycle/s, the impedance and current in the circuit will be respectively :

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 \[i=\frac{V}{2}=\frac{220}{50}=4.4\,A\]