question_answer

A 220 V, 50 Hz AC source is connected to an inductance of 0.2 H and resistance of 20 tl in series. The current in the circuit is

A)  3.33 A                                  

B)  33.3 A

C)  10 A                                     

D)  5 A

Correct Answer: A

Solution :

                 The inductive reactance\[{{X}_{L}}=\omega L=2\pi f\,L\] Given,   \[L=0.2H,\,\,f=50\,Hz\]                 \[\therefore \]  \[{{X}_{L}}=2\times \pi \times 50\times 0.2\]                 and        \[R=20\,\,\Omega \] Hence, impedance \[Z=\sqrt{{{X}_{L}}^{2}+{{R}^{2}}}\] \[\Rightarrow \]               \[Z=\sqrt{{{(2\pi \times 50\times 0.2)}^{2}}+{{(20)}^{2}}}\] and current \[I=\frac{V}{Z}=\frac{220}{\sqrt{3943.84+440}}\] \[\Rightarrow \]               \[I=\frac{220}{65.9}=3.33\,A\]