Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

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 ?

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

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

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):

\[Q=10volts x 10 x 10^{-6} farads=10^{-4}Coulombs\]

OpenStudy (anonymous):

O.K.Thank you radar.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!