A) \[\frac{{{R}_{1}}{{R}_{2}}{{Q}_{0}}}{2V}\]
B) \[\frac{2{{R}_{1}}{{R}_{2}}{{Q}_{0}}}{V}\]
C) \[\frac{{{R}_{1}}{{R}_{2}}{{Q}_{0}}}{\sqrt{2}V}\]
D) \[\frac{{{R}_{1}}{{R}_{2}}{{Q}_{0}}}{V}\]
Correct Answer: D
Solution :
[d] Current in coil 1 will attain its steady value \[{{I}_{01}}\frac{V}{{{R}_{1}}}\]infinite time. Let current in the two coils be \[{{i}_{1}}\] and \[{{i}_{2}}\]at time t. Using Kirchhoff s loop rule to the second coil we can write \[M\frac{d{{i}_{1}}}{dt}+{{L}_{2}}\frac{d{{i}_{2}}}{dt}-{{R}_{2}}{{i}_{2}}=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,Md{{i}_{1}}+{{L}_{2}}d{{i}_{2}}={{R}_{2}}{{i}_{2}}dt\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,M\int\limits_{0}^{{{I}_{01}}}{di+{{L}_{2}}}\int\limits_{0}^{{{I}_{02}}}{d{{i}_{2}}={{R}_{2}}}\int\limits_{t=0}^{\infty }{{{i}_{2}}dt}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,M{{I}_{01}}+{{L}_{2}}{{I}_{02}}={{R}_{2}}{{Q}_{0}}\] But \[{{I}_{02}}=0\] and \[{{I}_{01}}=\frac{V}{{{R}_{1}}}\] \[\therefore \,\,\,\,\,\,\,\,M\frac{V}{{{R}_{1}}}={{R}_{2}}{{q}_{0}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M=\frac{{{R}_{1}}{{R}_{2}}{{Q}_{0}}}{V}\]
You need to login to perform this action.
You will be redirected in
3 sec
