question_answer

A coil has an inductance of 0.7 H and is joined in series with a resistance of 220 f2. When the alternating emf of 220 V at 50 Hz is applied to it then the phase through which current lags behind the applied emf and the wattless component of current in the circuit will be respectively

A) \[~30{}^\circ ,\text{ }1\text{ }A\]

B) \[~45{}^\circ ,\text{ }0.5\text{ }A\]

C) \[~69{}^\circ ,\text{ }1.5\text{ }A\]     

D)  None of these

Correct Answer: B

Solution :

                 In an L-R circuit the phase angle is given by                 \[\phi ={{\tan }^{-1}}\frac{{{X}_{L}}}{R}\]                 where \[{{X}_{L}}\] is inductive reactance and R the resistance. Also,      \[{{X}_{L}}=\omega L=2\pi \,f\,L\] Given,   \[f=50\,Hz,\,\,L=0.7\,H\] \[\therefore \]  \[{{X}_{L}}=2\times 3.14\times 50\times 0.7\]                 \[=220\,\,\Omega \] and        \[R=220\,\,\Omega \] \[\therefore \]  \[\phi ={{\tan }^{-1}}\frac{220}{220}=1\] \[\Rightarrow \]               \[\phi ={{45}^{o}}\] Current \[I=\frac{E}{\sqrt{{{R}^{2}}+(\omega {{L}^{2}})}}\] Putting \[E=220\,V,\,\,\,\,R=220\,\,\Omega \],                 \[{{X}_{L}}=220\,\,\Omega \] \[\therefore \]  \[I=\frac{220}{\sqrt{{{(220)}^{2}}+{{(220)}^{2}}}}=\frac{1}{\sqrt{2}}\] Watt less current is \[I\] \[\sin \phi =\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=0.5\,A\]