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Algebraic Equation Manipulation for Electric Circuits
19 questions By Tony R. Kuphaldt
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Question 10 of 19
The decay of a variable over time in an RC or LR circuit follows this mathematical expression:
$$e^{-{\frac{t}{τ}}}$$
Where,
e = Euler’s constant ( ≈ 2.718281828)
t = Time, in seconds
τ = Time constant of circuit, in seconds
For example, if we were to evaluate this expression and arrive at a value of 0.398, we would know the variable in question has decayed from 100% to 39.8% over the period of time specified.
However, calculating the amount of time it takes for a decaying variable to reach a specified percentage is more difficult. We would have to manipulate the equation to solve for t, which is part of an exponent.
Show how the following equation could be algebraically manipulated to solve for t, where x is the number between 0 and 1 (inclusive) representing the percentage of original value for the variable in question:
$$x=e^{-{\frac{t}{τ}}}$$
Note: the “trick” here is how to isolate the exponent \(-\frac{-t}{τ}\). You will have to use the natural logarithm function!
Reveal answerShowing all the necessary steps:
$$x=e^{-{\frac{t}{τ}}}$$
$$In \ x = In - (e^{-{\frac{t}{τ}}})$$
$$In \ x = -\frac{t}{τ}$$
$$t = -τ \ In x$$
Notes:In my experience, most American high school graduates are extremely weak in logarithms. Apparently this is not taught very well at the high school level, which is a shame because logarithms are a powerful mathematical tool. You may find it necessary to explain to your students what a logarithm is, and exactly why it “un-does” the exponent.
When forced to give a quick presentation on logarithms, I usually start with a generic definition:
Given: ba = c Logarithm defined: logb c = a Verbally defined, the logarithm function asks us to find the power (a) of the base (b) that will yield c.
Next, I introduce the common logarithm. This, of course, is a logarithm with a base of 10. A few quick calculator exercises help students grasp what the common logarithm function is all about:
log10 = log100 = log1000 = log10000 = log100000 = log 1 10= log 1 100= log 1 1000= After this, I introduce the natural logarithm: a logarithm with a base of e (Euler’s constant):
Natural logarithm defined: lnx = loge x Have your students do this simple calculation on their calculators, and explain the result:
ln2.71828 = Next comes an exercise to help them understand how logarithms can ündo” exponentiation. Have your students calculate the following values:
e2 = e3 = e4 = Now, have them take the natural logarithms of each of those answers. They will find that they arrive at the original exponent values (2, 3, and 4, respectively). Write this relationship on the board as such for your students to view:
lne2 = 2 lne3 = 3 lne4 = 4 Ask your students to express this relationship in general form, using the variable x for the power instead of an actual number:
lnex = x It should now be apparent that the natural logarithm function has the ability to “undo” a power of e. Now it should be clear to your students why the given sequence of algebraic manipulations in the answer for this question is true.
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Question 11 of 19
Voltage and current gains, expressed in units of decibels, may be calculated as such:
AV(dB) = 10 log( AV(ratio) ) 2 AI(dB) = 10 log( AI(ratio) ) 2 Another way of writing this equation is like this:
AV(dB) = 20 logAV(ratio) AI(dB) = 20 logAI(ratio) What law of algebra allows us to simplify a logarithmic equation in this manner?
Reveal answerlogab = b loga Challenge question: knowing this algebraic law, solve for x in the following equation:
520 = 8x Notes:Logarithms are a confusing, but powerful, algebraic tool. In this example, we see how the logarithm of a power function is converted into a simple multiplication function.
The challenge question asks students to apply this relationship to an equation not containing logarithms at all. However, the fundamental rule of algebra is that you may perform any operation (including logarithms) to any equation so long as you apply it equally to both sides of the equation. Logarithms allow us to take an algebra problem such as this and simplify it significantly.
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Question 12 of 19
Solve for the value of x in the following equations:
10x = 80 x = 3 = 15 xx = Reveal answer10x = 80 x = 8 3 = 15 xx = 5 Notes:Have your students come to the front of the class and show everyone else the techniques they used to solve for the value of x in each equation. Remind them to document each and every step in the process, so that nothing is left to guess or to chance.
Question 14, Equation 2 - the ‘+’ symbol is missing between ‘a 3’.
Question 17, right-hand equation - the square-root should only be over the left side of the equation.
Question 19, Hint - the ‘+’ symbol is missing between ‘I 0.005’.
They do seem to be correct in the equivalent questions in the PDF version though.
In the answer to question 8, the second solution has R1 on the right side of the equation when it should be R.