Physics
WAEC 1991
Two capacitors C
1 and C
2 are connected as shown in the diagram. The capacitance C
2 is twice C
1 when the key is opened the energy stored up in C
1 is W. If the key is later closed and the system is allowed to attain electrical equilibrium, the total energy stored in the system will be
-
A.
\(\frac{1}{2}\)W
-
B.
\(\frac{2}{3}\)W
-
C.
W
-
D.
2w
-
E.
3W
Correct Answer: Option E
Explanation
E = \(\frac{1}{2} Cr^2 = \frac{1}{2}(C-1 + C_2)V^2 = (3C_1)V^2\)
= 3(\(\frac{1}{2}C_1v^2\)) = 3W
Report an Error
Ask A Question
Download App
Quick Questions
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}