That's right!UTF-8 U+6211 U+662F wrote:Agh! Well that sucks.mjcox2000 wrote:your third equation should have [math]I-\frac{V}{R_p}[/math], not [math]I-VR_p[/math], and that error messed up almost everything else you have. With that fix, you would have gotten part a right, and I haven't evaluated part b, but the way you approach it seems reasonable. Wolfram Alpha or Mathematica might help with the more involved integrals.a) [math]Q=CV[/math] [math]I_C=C \cdot \frac{dV}{dt}[/math] [math]I-\frac{V}{R_p} = C \cdot \frac{dV}{dt}[/math] [math]dt = C \cdot \frac{dV}{I-\frac{V}{R_p}}[/math] [math]t + C_{int} = -CR_p\ln(I-\frac{V}{R_p})[/math] (the constant of integration also being a C makes this slightly confusing) [math]t = -CR_p\ln(I-\frac{V}{R_p}) + k[/math] (so I changed it to a k) [math]t = -CR_p\ln(I-\frac{V}{R_p}) + CR_p\ln I[/math] (but plugging in (0,0) gives us a value for k) [math]\boxed{t_f = -CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I}[/math] b) I'm going to assume you mean energy for the next problem? [math]P = I^2R_s + \frac{V^2}{R_p}[/math] [math]E = I^2R_s \cdot (-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I) + \int^{t_f}_0\frac{V^2}{R_p}dt[/math] Now, let's solve for V in terms of t... [math]t = -CR_p\ln(I-\frac{V}{R_p}) + CR_p\ln I[/math] [math]\frac{CR_p\ln I - t}{CR_p} = \ln(I-\frac{V}{R_p})[/math] [math]\ln I - \frac{t}{CR_p} = \ln(I-\frac{V}{R_p})[/math] [math]Ie^{-\frac{t}{CR_p}} = I - \frac{V}{R_p}[/math] [math]V = IR_p(1-e^{-\frac{t}{CR_p}})[/math] [math]E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + \int^{t_f}_0\frac{(IR_p(1-e^{-\frac{t}{CR_p}}))^2}{R_p}dt[/math] [math]E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\int^{t_f}_0(1-e^{-\frac{t}{CR_p}})^2dt[/math] [math]E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left[t+\frac{e^{-2\frac{t}{CR_p}}-4e^{{-\frac{t}{CR_p}}}}{-\frac{2}{CR_p}}\right]^{t_f}_0[/math] [math]\boxed{E = CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left(-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I+\frac{\frac{(I-\frac{V_f}{R_p})^2}{I^2}-4\frac{I-\frac{V_f}{R_p}}{I}}{-\frac{2}{CR_p}}-\frac{3CR_p}{2}\right)}[/math] For part c: [math]\boxed{0 = \frac{d}{dV}\left[CI^2R_pR_s\ln\frac{I}{I-\frac{V_f}{R_p}} + I^2R_p\left(-CR_p\ln(I-\frac{V_f}{R_p}) + CR_p\ln I+\frac{\frac{(I-\frac{V_f}{R_p})^2}{I^2}-4\frac{I-\frac{V_f}{R_p}}{I}}{-\frac{2}{CR_p}}-\frac{3CR_p}{2}\right)\right]}[/math] Although you'd need to test all of the roots and see which one yields the lowest value

Instead of integrating for [math]V_{capacitor}(t)[/math], you can eliminate [math]R_s[/math] from the circuit since it impacts just the power loss and not the charging time. Then, [math]I[/math] is in parallel with [math]R_p[/math], and you can use Norton-Thevenin equivalence to see that the capacitor is simply charged by a voltage source [math]IR_p[/math] through resistance [math]R_p[/math], so you can apply a stock formula rather than integrating. Your part b formula is right, but it can be algebraically simplified to [math]-\frac12 c \bigg[V_f (2 I R_p + V_f) + 2 I^2 R_p (R_p + R_s) ln\bigg(1 - \frac{V_f}{iR_p}\bigg)\bigg].[/math] I haven't proved it, but intuition tells me there can only be one point at which the derivative of part b is 0. If current is under [math]V_f/R_p[/math], infinite energy is wasted because the target voltage is never reached. With current slightly above that threshold, it takes a very long time to charge, so a lot of energy is wasted in the series resistor. As current increases, time decreases, which should monotonically decrease the energy wasted in the series resistor. However, when current is very high, the parallel resistor wastes a lot of energy because it shorts around the capacitor. If I'm not mistaken, the limiting energy waste for small and large currents mean that there should be just one local minimum. (Plots I've done seem to confirm that.)