A) Zero
B) \[2L\]
C) \[\frac{L}{2}\]
D) \[L\]
Correct Answer: C
Solution :
When two coils\[{{L}_{1}}\]and\[{{L}_{2}}\]are connected in parallel between two points and current\[i\]is divided between them, then \[i={{i}_{1}}+{{i}_{2}}\] \[\frac{\Delta i}{\Delta t}=\frac{\Delta {{i}_{1}}}{\Delta t}+\frac{\Delta {{i}_{2}}}{\Delta t}\] When currents through inductances are growing, induced emf?s are set up in them than, \[e=-{{L}_{1}}\frac{\Delta {{i}_{1}}}{\Delta t}=-{{L}_{2}}\frac{\Delta {{i}_{2}}}{\Delta t}\] If L be the equivalent resistance, then \[e=-L\frac{\Delta i}{\Delta t}\] \[\Rightarrow \] \[\frac{e}{L}=-\frac{\Delta i}{\Delta i}=-\left( \frac{\Delta {{i}_{1}}}{\Delta t}+\frac{\Delta {{i}_{2}}}{\Delta t} \right)=\frac{e}{{{L}_{1}}}+\frac{e}{{{L}_{2}}}\] \[\Rightarrow \] \[\frac{1}{L}=\frac{1}{{{L}_{1}}}+\frac{1}{{{L}_{2}}}\] \[\Rightarrow \] \[L=\frac{{{L}_{1}}{{L}_{2}}}{{{L}_{1}}+{{L}_{2}}}\] Given, \[{{L}_{1}}={{L}_{2}}=L\] \[\therefore \] \[L=\frac{{{L}^{2}}}{2L}=\frac{L}{2}\]
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