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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.

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For convenience write RC=D and solve the ODE dV/dt=-V/D which gives 
V=Ae^(-t/D) by separating the variables. 
When t=0, V=10 and this gives A=10, 
when t=2, V=1 giving 1=10e^(-2/D) and you can find D by taking lns of 
each side. D=RC, so R=D/C 
You should get 2.90X10^5 ohms

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