Circuit Diagnosis of Prioritization of the Main Ideas

The instruction was to turn each of the following circuit diagrams into a first or second order linear differential equation in standard form.

1. A resistor in series with an inductor and a power supply, zero current at t = 0.2. A resistor in series with a capacitor and a power supply, zero charge at t = 0.3. An inductor in series with a capacitor, full charge charge Q0 at t = 0.4. A resistor in series with an inductor, lots of current at t = 0.5. A resistor in series with a capacitor lots of charge at t = 0.6. An inductor in series with a capacitor and a power supply.

Only number 6 leads to math that is beyond the simple differential equation solutions of this class. Allowing that ε is a constant and represents power supply voltage, the other 5’s differential equations are:

1. dI/dt + (R/L)I = ε2. dQ/dt + (1/RC)Q = ε3. d2Q/dt2 + (1/LC)Q = 04. dI/dt + (R/L)I = 05. dQ/dt + (1/RC)Q = 0

The instruction was to write the solution for each of the above for its time-dependent variable. That’s either I(t) or Q(t) depending on the situation. A person should at this point have little trouble realizing that there are roughly three ways to confirm each of the following solution forms:

1. I(t) = IT(1 – e – t/τ) where IT = ε/R and τ = L/R

If these 5 are memorized instead of quickly proven on a case-by-case basis every time, then you are wasting your time, and you’ll find out why if you don’t know already. It’s the whole point of this document. And someone who says that these five solutions are time-consuming, because they are separation of variables continues to miss the point.

2. Q(t) = QT(1 – e – t/τ) where QT = Cε and τ = RC3. Q(t) = Q0cost) where ω = (LC) – ½ 4. I(t) = I0e – t/τ where τ = L/R5. Q(t) = Q0e – t/τ where τ = RC

And manipulations of the above are possible. For example, if you wanted capacitor voltage instead, you’d multiply Q(t) by (1/C). Or if you wanted inductor voltage, you’d take I(t)’s derivative and then perform the product L(dI/dt), and L(dI/dt) would be a function of t.

Now you go to the next page and see if you understand what’s important.

Diagnostic Problem 1 In a circuit that has a capacitor C in series with a resistor R1 and a power supply VB, and at t = 0 the capacitor is uncharged,

the loop rule applies in the form: VB = R1I + Q/C, and this leads to a differential equation that is solved simply and produces the following time-dependent formula for the current that runs through R1:

I(t) = (VB/R1)e-t/τ where τ = R1C

Don’t take my word for it. Do a quick proof that solves for Q(t), and then take Q’s derivative to get I(t). If you don’t end up with the expression above, something went wrong.

The following 5 quick things are not mastery level and take 20 seconds each:

1. Find the current when time equals 2τ if the battery voltage is 9 V and the R1 value is 10 Ω.2. Is the current a maximum at the instant when the power is turned on or is it zero?3. Is VB/R1 the value of current initially, after a really long time, or neither?4. After a really long time, is the capacitor’s charge QT or is it zero or neither?5. Find the value of QT if τ = 2 seconds.

Answers: VB/R1 = 0.9 A, so the first answer is (0.9 A)e – 2 , because that’s the communication that τ is for. People who go and plug in R and C to solve for τ in s don’t know this, and they do things the hard way. No value of s ever had to be determined. For #5, QT = (1/C)VT = (0.9 V)/C but C was never given, so at some point it had to be exploited that C = τ/R1. But none of those 5 items were the main idea.

6. In the circuit shown below, the capacitor is uncharged at t = 0, and t = 0 is the instant when the battery is first turned on. R2 will be our name for the 15 Ω resistor. R1 will be our name for the 10 Ω resistor. At the instant t = 0.5 s, the current through R2 is 0.42 A.

A. Find the current through R1 at that instant.B. Find the charge on the capacitor at that instant.

Part 3:

Suppose in the picture above, the capacitor becomes fully charged after a very long time. Then the left section containing the power supply is disconnected somehow.

7. Find how long it will take for the capacitor to reduce to 37% of the maximum charge that it had.

8. Find the amount of energy stored in the capacitor at the instant solved for in #1.

Answers on the following pages. Do all problems before you look. And if you take longer than 5 minutes to do questions 6 through 8, you are missing the point. So you stop, then read the key, and diagnose yourself because you have to adjust if you want to move forward always stressing the main ideas and not wasting time with whatever complicating distractions you stressed in the first attempt.

Diagnosis Example 1 Answers

1. 0.9 A times e-2

2. Maximum. Also, its starting value is 0.9 A.3. See #24. QF

5. QF = CVF = CVB = (τ/R1)VB = (VB/R1)τ = (0.9 C/s)(2 s) = 1.8 CAll of the above way simple. Now come the life lessons:

6. New circuit to solve for: after looking at the circuit and noticing that there are three different resistances, suppose that someone spent all their time wondering what values of R’s they should use in the formula

I(t) = (VB/R)e-t/τ where τ = RC

I suppose this person could think there are many options for the R’s:

10 Ω since that is R1

6 Ω since that is the equivalent of parallel combo of the 15 Ω and the R1 when no capacitor is there.

13 Ω since that is what would be thought if R1 and the 3 Ω were considered in series

9 Ω since that would be the resistance of the whole thing if the capacitor were not there.

And maybe this person would spend time wondering if the R in τ expression and the R in the I(t) expression are even the same R.

This person would be entirely wasting all of his/her thinking time, because their way of looking at math is entirely wrong. The I(t) = (VB/R)e-t/τ formula is 100% irrelevant, because it doesn’t apply, because it was derived only from a loop rule expression that involved three specific members in a series. That 3-Member series NEVER EXISTS in problem 6. Anyone who doesn’t know why doesn’t know what a series is. Series circuit elements have one common current. The #6 circuit has three separate current values at any one time, and its differential equation that would lead to something like I(t) = (some constant)e-t/(some other constant) would not nearly be as simple, and would certainly not a be a proof that has been done on these pages. That’s why I train you to know where the simple differential equations come from, as stressed on Page 1. The idea is for you to know that the lack of proof allows you to say that I(t) = (some constant)e-t/(τ) is NOT the heart of the solution and to instead look at the fundamentals that are Kirchhoff #1 and Kirchhoff #2 and not be distracted by irrelevancies. (By the way, there is a true solution that contains a e-t/(τ) term, but I in my head do not know what τ would be structured as. Again, you can understand this if you did Page 1 by proving and not by memorization.)

To belabor the point, quoting myself on Page 2, “In a circuit that has a capacitor C in series with a resistor R1 and a power supply, the loop rule applies in the form: VB = R1I + Q/C, and this leads to a differential equation that is solved simply and produces the following time-dependent formula for the current that runs through R1: I(t) = (VB/R)e-t/τ “

If one isn’t distracted, then in solving #6, anything related to deriving I(t) = (?)e-t/τ is irrelevant.

The true solution is a measure of whether the student knows Kirchhoff’s Laws. Those have been repeatedly described as the central concepts. Real solution occurs now:

Solution:

0.42 A flows through R2. The voltage across R2 is therefore (0.42 A)(15 Ω) = 6.3 V. I have no idea what to do from here, so I express loop and branch equations as I’ve been trained to. Just like FBD’s and Fnet = ma, these expressions are central and will solve themselves if I am aware and trained. STOP: don’t look at the formula expressions that are about to give this away. That doesn’t test your understanding. Take this paragraph you just read and DO WHAT I SAY in it by writing it in equation form. Don’t copy me doing equations. YOU build the equations, a loop rule equation for each loop, a current addition equation for each junction. The whole point is that you have the experience expressing Kirchhoff’s Rule 1 and Kirchhoff’s Rule 2. Don’t watch me do it again. Do it yourself and check the answer, and don’t read my solution until you are done.

When you are done, the answers are:

0.48 A for part a 0.3 C for part b

Solution now given away if one still hasn’t done it:

So simple that if you can’t do it in paragraph form before math form, something is missing. The following paragraph solution is better than the one that resorts to equation forms. The clear verbal solution, like that in quotes, comes first, and it’s translated into equations afterwards. When equations are presented before the problem is conceptually understood, failure every time.

Part a: “Since R2I = 6.3 V are across R2, then 2.7 V must be across 3 Ω, because that is all that is allowed by the three member loop that contains R2, the 3 Ω thing, and the 9 V PS. Well, anyone who knows Ohm’s Law knows that 2.7 V across the 3 Ω thing yields (2.7÷3) A of current, 0.9 A. At the junction, these 0.9 A of current split into the 0.42 A given in the middle and the (0.9 A – 0.42 A) that go the other way. Answer: 0.48 A.” And this entire statement is translated into math language as equations 1 and 3 below.

Part b: “If 0.48 A flow through R1, the voltage across it is R1I = 4.8 V. But we already know that 6.3 V are across the R1 and C combined. We know this from applying the loop rule to the right loop. So Vcap = 1.5 V, the difference between 6.3 and 4.8.” This statement is translated into math language as equation 2 below. Also, it was really easy to do in my head.

Solution Equations:

Doing algebra on what’s below has to be seen as equivalent to the verbal quotes above. If not, don’t ever bother writing a formula as your solutions statement. It will fail.

Eqn. 1: 9 V – (3 Ω)Ibefore split – (6.3 V) = 0 That’s a real simple loop on the left. (K2)

Eqn. 2: R2(0.42 A) = Vcap + (10 Ω)Ithrough R1 That’s a different simple loop. (K2)

Eqn. 3: Ibefore split = 0.42 A + Ithrough R1 That’s conservation of charge (K1 junction rule)

But 1.5 V is not the answer to b. It asks for charge. I have to know the definition of capacitance:

Q = CVcap = (0.2 F)(1.5 V) = 0.3 C

Diagnostic Problem 2:

We know that when we have an inductor L in series with a resistor R and a constant power supply voltage ε, that if it is connected at t = 0, the current will obey the function:

I(t) = IT(1 – e – t/τ) where IT = ε/R and τ = L/R

If the following circuit is assembled and connected at t = 0, solve at the instant shown:

In case you can’t read it, the inductor has 50 H. The vertically drawn resistance above it has 5 Ω.

A. Solve for the current that runs in the inductorB. Solve for the value of the derivative dI/dt, where I refers to the current in the inductor.C. Using the knowledge of Parts A and B to form the structure, draw the shape of the full

graph of I vs. t for time values that go from t = 0 to infinity. You’ll need to consider whether I goes up or down as time goes on and then what happens to its slope as I’s magnitude changes. It’s difficult to be numerical on the t axis of this graph, but the I axis needs clear values marked for the starting value of I and the eventual value of I.

Answers:

A. 1.65 AB. 0.142 A/s

C. Current is rising. As it does, the voltage across the 5 Ω thing goes up (because ΔV = IR) which brings the inductor voltage down (by the loop rule). The lower the inductor voltage, the lower dI/dt is (by Faraday’s Law). So the higher the graph goes, the less slope it has. So this is a rising, concave down curve. Its starting value of A and its asymptotic final value of A must be solved for and specified, and neither of them are zero. I’ll let you do that. If you don’t, don’t ask about, and eventually claim you don’t know how, you’ll end up failing some graded thing. I’m saying it loud and clear that I’ll answer the question only if you ask. If you don’t ask, the requirement is that you solve for those two A values on your own. This is supposed to become do it for yourself, not for me, so I’m not spoon-feeding every answer. I’ll solve anything you ask.

An Indication that the main idea is completely missed:

If the solver starts by trying to answer Questions A and B by writing:

I(t) = IT(1 – e – t/τ)

or some form similar to it.

Good luck with that vague formula that has unspecified constants and total meaninglessness for the circuit at hand. Does a person write that math form, thinking it will yield partial credit, without knowing exactly what I(t) = IT(1 – e – t/τ) would mean if it did apply? No, it yields zero partial credit, and it deserves negative credit, because it says, “I’m not going to think about the physics right now, but I’m gonna see if I can cheat myself by writing something meaningless.”

By the same explanation as on Pages 4 and 5, the form I(t) = IT(1 – e – t/τ) is of zero help on this problem, because no proof has been shown for how to model τ. But why would that distract anyone? This is the second simplest* Kirchhoff Problem around. No proof will be offered here that solves a differential equation (which would be complicated), because that distracts from what this problem really is:

Solution: “17 V across the 20 Ω resistor. This is by Ohm’s Law. The Loop Rule then dictates 25 V

across the 10 Ω resistor. Ohm’s Law then dictates a current of 2.5 A through the 10 Ω resistor.” From here, the rest of the problem is simple for parts A and B. Do it yourself, because the answers to prove are at the top of this page. If you don’t ask about these in class, it means you plan to be self-sufficient and linguistic with Kirchhoff’s Laws every time just like the quotation part of this paragraph is modeling. If you have no intention of being linguistic, you aren’t learning circuits. I am hereby offering to answer solving-oriented questions on circuit language.

*The simplest contain no junctions. This one contains 1 junction.