| An inductor of inductance \[L=400\text{ }mH\]and resistors of resistances \[{{R}_{1}}=2\text{ }\Omega \] and \[{{R}_{2}}=2\Omega \] are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is: [AIEEE 2009] |
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A) \[6\text{ }{{e}^{5t}}V\]
B) \[\frac{12}{t}{{e}^{-3t}}V\]
C) \[6\left( 1-{{e}^{\frac{-t}{0.2}}} \right)V\]
D) \[12{{e}^{-5t}}V\]
Correct Answer: D
Solution :
| [d] \[{{e}_{L}}=E{{e}^{\frac{t{{R}_{2}}}{L}}}\] |
| \[{{e}_{L}}=12{{e}^{-\frac{2}{.4}t}}\] |
| \[{{e}_{L}}=12{{e}^{5t}}.\] |
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