A) \[0.1\times {{10}^{-4}}s\]
B) \[0.2\times {{10}^{-4}}s\]
C) \[0.3\times {{10}^{-4}}s\]
D) \[0.4\times {{10}^{-4}}s\]
Correct Answer: C
Solution :
Given, \[L=1.8\,\times {{10}^{-4}}\,H,\,\,R=6\Omega \] Self-inductance of each coil \[=\frac{1.8\,\times {{10}^{-4}}}{2}H\] Resistance of each coil \[=\frac{6\Omega }{2}=3\Omega \] Coils are then connected in parallel \[\therefore \,\,L'=\frac{\frac{1.8}{2}\times {{10}^{-4}}\,\times \frac{1.8}{2}\times {{10}^{-4}}}{\frac{1.8}{2}\times {{10}^{-4}}+\,\frac{1.8}{2}\,\times {{10}^{-4}}}\] \[=0.45\,\times \,{{10}^{-4}}H\]and \[R'\,=\frac{3\times 3}{3+3}\,=1.5\Omega \]time constant \[\frac{L'}{R'}\,=\frac{0.45\,\times {{10}^{-4}}}{1.5}\] \[=0.3\,\times {{10}^{-4}}s\]You need to login to perform this action.
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