question_answer

 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  Solved  Paper-2009

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 :

\[{{e}_{L}}=E{{e}^{\frac{t{{R}_{2}}}{L}}}\]                 \[{{e}_{L}}=12{{e}^{\frac{2}{.4}t}}\]                 \[{{e}_{L}}=12{{e}^{5t}}.\] Directions: Question numbers 28, 29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram.