A) 1/3
B) 2/3
C) 1
D) 4/3
Correct Answer: B
Solution :
The given, \[L=4.0mH=4.0\times {{10}^{-3}}H\] \[{{X}_{L}}=\omega L=2\pi fL\] \[=2\times 3.14\times 50\times 4.0\times {{10}^{-3}}\Omega \] \[=1256\times {{10}^{-3}}\Omega \] \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] \[=\sqrt{{{9}^{2}}+{{(1.26)}^{2}}}\] \[=\sqrt{82.5876}\] \[=9.08\,\Omega \cong 9.0\,\Omega \] So, total resistance \[=9+9+9=27\Omega \] According to the question, \[{{i}_{1}}=\frac{E}{Total\,resis\tan ce}=\frac{18}{27}\] ?. (i) \[{{i}_{2}}=\frac{18}{18}=1\] The ratio of the currents \[{{i}_{1}}:{{i}_{2}}=\frac{18}{27}:1=2:3\]You need to login to perform this action.
You will be redirected in
3 sec