When a charged capacitor is connected to earth, it discharges and the voltage V across it will decrease according to the equation:
dV/dt=−V/RC
where R (the resistance) and C (the capacitance) are constants. If the voltage V across a certain capacitor with capacitance C = 3×10−6 farads
drops from 10 volts to 1 volt in two seconds, determine the resistance R.

Question

When a charged capacitor is connected to earth, it discharges and the voltage V across it will decrease according to the equation:
dV/dt=−V/RC
where R (the resistance) and C (the capacitance) are constants. If the voltage V across a certain capacitor with capacitance C = 3×10−6 farads
drops from 10 volts to 1 volt in two seconds, determine the resistance R.

anonymous · Answer

dv/dt means a small change in v/ small change in time right? what physics class is this for (so I can see what method you want to use)

anonymous · Answer

calculus actually

anonymous · Answer

ideally you would solve this as a differential equation. dV/V=dt/(-RC)
integrate both sides and you get ln(V)=t/(-RC). Take -RC to the other side and plug in your values.

anonymous · Answer

whoops missing a part. you want to solve for R, so then the final equation is R=t/C*ln(V)

anonymous · Answer

does this make sense?

anonymous · Answer

i got 1464816 ohms. does that seem right? :/

anonymous · Answer

i mean 303413.0755

anonymous · Answer

looks right to me. does the calculus make sense to you?

anonymous · Answer

yes thank you