A) \[\frac{4}{\sqrt{7}}A\]
B) \[1.0A\]
C) \[\frac{4}{7}A\]
D) \[0.8A\]
Correct Answer: D
Solution :
\[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] Here, \[R=4\Omega ,{{X}_{L}}=L\omega =3\times {{10}^{-3}}\times 1000\Omega \] \[=3\Omega \] Then, \[Z=\sqrt{{{(4)}^{2}}+{{(3)}^{2}}}\] \[Z=\sqrt{16+9}=\sqrt{25}\] \[Z=5\,\Omega \] Hence, current \[{{I}_{0}}=\frac{{{E}_{0}}}{z}=\frac{4}{5}=0.8\,A\]
You need to login to perform this action.
You will be redirected in
3 sec
