question_answer

Figure shows a circuit that contains three identical resistors with resistance \[R=9.0\,\Omega \], two identical inductors with inductance L = 4.0 mH, and a battery with emf E = 18 V. Find the ratio of the currents in the circuit just after and long after the switch K is closed.

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                 \[{{i}_{1}}:{{i}_{2}}=\frac{18}{27}:1=2:3\]