A) \[\frac{V{{R}_{1}}{{R}_{2}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}\] at t=0 and \[\frac{V}{{{R}_{2}}}\] at \[t=\infty \]
B) \[\frac{V}{{{R}_{2}}}\] at t=0 and \[\frac{V({{R}_{1}}+{{R}_{2}})}{{{R}_{1}}{{R}_{2}}}\] at \[t=\infty \]
C) \[\frac{V}{{{R}_{2}}}\] at t=0 and \[\frac{V{{R}_{1}}{{R}_{2}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}\] at \[t=\infty \]
D) \[\frac{V({{R}_{1}}+{{R}_{2}})}{{{R}_{1}}{{R}_{2}}}\] at t=0 and \[\frac{V}{{{R}_{2}}}\] at \[t=\infty \]
Correct Answer: C
Solution :
[c] At t = 0, no current will flow through L and \[{{R}_{1}}\] \[\therefore \] Current through battery \[=\,\frac{V}{{{R}_{2}}}\] At \[t=\infty \] effective resistance, \[{{R}_{eff}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] \[\therefore \] Current through battery\[=\frac{V}{{{R}_{eff}}}=\frac{V\,({{R}_{1}}{{R}_{2}})}{({{R}_{1}}+{{R}_{2}})}\]You need to login to perform this action.
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