Southtown 30,651 Posted August 9, 2011 Report Share Posted August 9, 2011 I could use some help interpreting this calculus equation into something more trig level.From Hyperphysics.phy-astr.gsu.edu I assume this translates to V_{b} = Q/(C(1/1/(t/RC))) (I need to find V_{b}) If someone could just arrange this equation into something I can understand and explain I would greatly appreciate it. A college textbook for electronics 101 asks for the voltage across a 15nF capacitor charged with 25 volts 25ms after its leads are shorted with a 2MΩ resistor, but the book never touches on that kind of math. Go figure. It only shows how to get the time constant (t_{c} = resistance x capacitance) plus the five-increment charge/decay chart. (0%)/63.2%/86.5%/95%/98.2%/99.3% Charge(100%)/36.8%/13.5%/5.0%/1.8% DecayWouldn't really be a problem if the time constant, 30ms, was a clean multiple of the time given, 25ms (which is usually the case in this book.) I wonder if they meant 12.5nF instead of 15. PS the book cost 137.50 USD <_< [/finger] Quote Link to post Share on other sites

lawcat 95,363 Posted August 9, 2011 Report Share Posted August 9, 2011 I don't use mathlab symbols but I'll try to explain it anyway. The cap is charged and shorted, meaning you have no source in the circuit other than the capacitor voltage. When you short the capacitor's leads, all you have in the circuit is the capacitor and the resistor in series. The capacitor discharges current through the resistor. Using Kirchoff's current law at the node between the cap and R, the current from cap "into" node (+) = current "leaving" the node (-). The current from capacitor is I=CdV/dt. Current into resistor is V/R. So the formula is: CdV/dt = - V/R; or CdV/dt + V/R = 0. Or rearranged: dV/dT + (1/RC)V = 0. This is first order linear diff using "e ^ Integral of p(t)dt" where p(t)dt = 1/CR to multiply both sides with. (e^Integral(1/RC)dt)dV/dT = 0 Now to get V you integrate both side, keeping in mind that where the 0 is, that is where the initial condition is imposed of the voltage at 25 V. So when you integrate 0 on the right you will ebe left with 25 V. So after integration on both sides with respect to t: (e^t/RC)(V(t))=25 V or V(t)=25V x e^-t/RC plug and chug. Southtown 1 Quote Link to post Share on other sites

Southtown 30,651 Posted August 9, 2011 Author Report Share Posted August 9, 2011 Thanks for the help, man. I'll try that out. Quote Link to post Share on other sites

Southtown 30,651 Posted August 10, 2011 Author Report Share Posted August 10, 2011 I'm assuming the 'e' just means: x 10^^{x}. Guess that's been my trouble from the beginning. -t/RC = -0.025s/(2MΩ·15nF) = -0.025s/0.03s = -0.833 V(t) = 25v e -0.833 But I can't get any calculator to put decimal points in the exponent. Good news is I figured out the online calc at hyperphysics lol Got 10.43v there but still just trying to figure out how. PS 6.847e-2 maybe? Quote Link to post Share on other sites

lawcat 95,363 Posted August 10, 2011 Report Share Posted August 10, 2011 V(t) = 25v e -0.833Got 10.43vYup, that is the answer, 10.86V remain at capacitor after 25ms. there but still just trying to figure out how. 1. When you are trying to find Voltage at a point in circuit, you apply Kirchoff's current law at the node where you are trying to find voltage.2. You get some equation. For purely resistive circuits (nocaps or inductors), you will get some equation that is usually solved by matrix. You'll have to learn how to solve matrices.3. If the circuit contains caps or inductors, you will end up with a differential equation, and you'll have to learn how to solve differential equations.4. Luckily, in electronics you will often be solving partial circuits involving RC, LC, RL, or RLC. Formula's stemming from differential equations have been solved for these circuits long time ago. You'll just have to learn them and their application.5. Here, for an RC circuit in series, all you need to do is remember that the voltage between cap and R in RC series circuit is V = Vinitial x e ^ - t / RC. The key is to recognize the type of circuit you have and know which formula applies. PS 6.847e-2 maybe?no. Quote Link to post Share on other sites

Southtown 30,651 Posted August 10, 2011 Author Report Share Posted August 10, 2011 1. When you are trying to find Voltage at a point in circuit, you apply Kirchoff's current law at the node where you are trying to find voltage.2. You get some equation. For purely resistive circuits (nocaps or inductors), you will get some equation that is usually solved by matrix. You'll have to learn how to solve matrices.3. If the circuit contains caps or inductors, you will end up with a differential equation, and you'll have to learn how to solve differential equations.4. Luckily, in electronics you will often be solving partial circuits involving RC, LC, RL, or RLC. Formula's stemming from differential equations have been solved for these circuits long time ago. You'll just have to learn them and their application.5. Here, for an RC circuit in series, all you need to do is remember that the voltage between cap and R in RC series circuit is V = Vinitial x e ^ - t / RC. The key is to recognize the type of circuit you have and know which formula applies.Thank you very much. Lots to go on. Thing is this question was obviously a typo considering the scope of the course. I'll just include this bit in my answer (as opposed to cheating via hyperphysics): V_{c} = V_{o}e^{-t/RC}V_{c} = <SYNTAX ERROR>V_{c} = Abort, Retry, Ignore Edit: Talked to the instructor, and now I feel stupid. lol The e is epsilon ε and it's just a button on my calculator. Yay Thanks again, lawcat. Quote Link to post Share on other sites

Qfwfq 95,327 Posted August 15, 2011 Report Share Posted August 15, 2011 I'm assuming the 'e' just means: x 10^^{x}.No, [imath]e[/imath] is the base for natural logarithms and raising [imath]e[/imath] to a power: [imath]e^x[/imath] is the inverse procedure of [imath]\ln x[/imath] (taking the natural log of [imath]x[/imath]). Therefore, if you can find the [imath]\ln[/imath] key on a scientific calculator and you also know how to invert its functions (just like using the [imath]\sin[/imath] key to calulate an [imath]\arcsin[/imath] for instance) then you can also compute the exponential that you want. In a similar manner you can also check that inverting the [imath]\log[/imath] (common or vase ten logarithm) gives you a comutation of [imath]10^x[/imath]. Apart from this, the [imath]V_b[/imath] in that formula stands for the voltage at time [imath]t=0[/imath] which is a given in the exercise, not something you need to compute. I don't know what to say about that books shortcomings, in ratio to its price. Quote Link to post Share on other sites

Southtown 30,651 Posted January 19, 2012 Author Report Share Posted January 19, 2012 No, [imath]e[/imath] is the base for natural logarithms and raising [imath]e[/imath] to a power: [imath]e^x[/imath] is the inverse procedure of [imath]\ln x[/imath] (taking the natural log of [imath]x[/imath]). Therefore, if you can find the [imath]\ln[/imath] key on a scientific calculator and you also know how to invert its functions (just like using the [imath]\sin[/imath] key to calulate an [imath]\arcsin[/imath] for instance) then you can also compute the exponential that you want. In a similar manner you can also check that inverting the [imath]\log[/imath] (common or vase ten logarithm) gives you a comutation of [imath]10^x[/imath]. Apart from this, the [imath]V_b[/imath] in that formula stands for the voltage at time [imath]t=0[/imath] which is a given in the exercise, not something you need to compute.Thanks Q, ya my teacher showed the [imath]e[/imath] key. lol ~> felt pretty stupidI misunderstood that in the equation as engineering notation. I don't know what to say about that books shortcomings, in ratio to its price.Haha, ya it's gotta be a scam. Quote Link to post Share on other sites

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