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.You need to login to perform this action.
You will be redirected in
3 sec