CapacitanceHard
Question
A 2μF capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is
Options
A.0 %
B.20 %
C.75 %
D.80 %
Solution
Ui = 1/2× 2 × V2 = V2
qi = 2V
Now, switch S is turned to position 2
qi = 2V
8V - 4q = q
⇒
ᐃH = V2
qi = 2V
Now, switch S is turned to position 2
qi = 2V
8V - 4q = q
⇒
ᐃH = V2
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