A) \[6{{e}^{-5t}}V\]
B) \[\frac{12}{t}{{e}^{-3t}}V\]
C) \[6(1-{{e}^{-t/0.2}})V\]
D) \[12{{e}^{-5t}}V\]
Correct Answer: D
Solution :
[d]: If \[{{I}_{1}}\]is the current through \[{{R}_{1}}\] and \[{{I}_{2}}\]is the current through L and \[{{R}_{2}},\]then \[{{I}_{1}}=\frac{\varepsilon} {{{R}_{1}}}\]and \[{{I}_{2}}={{I}_{0}}(1-{{e}^{-t/\tau }}),\] Where \[\tau =\frac{L}{{{R}_{2}}}=\frac{400\times {{10}^{-3}}}{2}=0.2s\] and\[{{I}_{0}}=\frac{\varepsilon }{{{R}_{2}}}=\frac{12}{2}=6A\] Thus,\[{{I}_{2}}=6(1-{{e}^{-t/0.2}})\] Potential drop across L, i.e., \[\varepsilon -{{R}_{2}}{{I}_{2}}=12V-2\times 6(1-{{e}^{-t/0.2}})V=(12{{e}^{-5t}})V\]
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