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Integration question

When connected to earth, any charged capacitor will discharge, and the voltage across the capacitor will decrease. This is modelled by the differential equation dV/dT=-V/RC, where V = voltage, and R and C (resistance in ohms and capacitance in farads respectively) are constants. If V drops from 10 volts to 1 volt in two seconds, and is C = 3*10^-6 farads, determine the value of R.
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1 Answer

dV/dt = -V / RC
 
dV/V = -dt / RC
 
V(t) = V(0)e-t / RC
 
C = 3 x 10-6 F
V(0) = 10
V(2) = 1
 
V(2) / V(0) = e-2 / RC
 
1/10 = e-2 / RC = 0.1
 
-2 / RC = ln(0.1)
 
R = -2 / [C ln(0.1)] = -2 / [(3 x 10-6 F) ln(0.1)] ≅ 289,530 Ω 
 
 
 

Comments

I am using the given initial and final conditions for V(t) on the time interval [0,2] to solve for the unknown R.  The general solution to the differential equation gives the exponential form of the function, with an unknown constant and an unknown R.  The condition at t = 0 determines the constant.  Once you now that constant, you can find R by applying the condition at t = 2.  
 
And you are welcome, Andy.