How long will it take to discharge a 10μF capacitor from 10V to zero if the current through the capacitor is constant and equal to 2A ?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
q=Cv
OpenStudy (anonymous):
Thank you.
OpenStudy (anonymous):
something along those lines
OpenStudy (anonymous):
wait that question is dodgy
OpenStudy (anonymous):
current cant be constant through a capacitor
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
wait , take back what I said
OpenStudy (anonymous):
\[i= c \frac{dv}{dt}\]
OpenStudy (anonymous):
\[c \frac{dv}{dt} = 2 \]
OpenStudy (anonymous):
divide both sides by c , integrate, use the initial condition to find the constant of integration
OpenStudy (anonymous):
etc
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (radar):
2 amperes represent 2 coulombs/sec. Since there is only 10^-4 coulomb of charge, if the currents was constant, the capacitor would be completely discharged in 50 micro seconds.
OpenStudy (anonymous):
\[v = \frac{2t}{c} +D\]
where D is integration constant, which is 10 from the initial condition
OpenStudy (anonymous):
O.K.Thank you.
OpenStudy (radar):
The amount of charge was computed from the equation provided by elecengr above.
Q=CV where Q is charge in Coulombs, C is Capacitancer in farads, v voltage in volts.
OpenStudy (radar):
\[Q=10volts x 10 x 10^{-6} farads=10^{-4}Coulombs\]
Still Need Help?
Join the QuestionCove community and study together with friends!