A fully charged 6.7 uF capacitor is connected in series with a 1.7×10^5 Ohm resistor. A....
A. What percentage of the original charge is left on the capacitor after 1.7 s of discharging?
Homework Answers
Q = Qo (e-t/RC) t = 1.7 , R = 1.7*10^5 , C = 6.7 *10^-6
on putting we get
Q/Qo = 0.2248
percentage = 22.48 percent os charge is still left
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A fully charged 6.7 uF capacitor is connected in series with a
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