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Algebraic Substitution for Electric Circuits
13 questions By Tony R. Kuphaldt
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Question 1 of 13
There are two basic Ohm’s Law equations: one relating voltage, current, and resistance; and the other relating voltage, current, and power (the latter equation is sometimes known as Joule’s Law rather than Ohm’s Law):
E = I R P = I E In electronics textbooks and reference books, you will find twelve different variations of these two equations, one solving for each variable in terms of a unique pair of two other variables. However, you need not memorize all twelve equations if you have the ability to algebraically manipulate the two simple equations shown above.
Demonstrate how algebra is used to derive the ten öther” forms of the two Ohm’s Law / Joule’s Law equations shown here.
Reveal answerI won’t show you how to do the algebraic manipulations, but I will show you the ten other equations. First, those equations that may be derived strictly from E = I R:
$$I = \frac{E}{R}$$
$$R = \frac {E}{I}$$
Next, those equations that may be derived strictly from P = I E:
$$I = \frac {P}{E}$$
$$E = \frac {P}{I}$$
Next, those equations that may be derived by using algebraic substitution between the original two equations given in the question:
$$P = I^2R$$
$$P = \frac{E^2}{R}$$
And finally, those equations which may be derived from manipulating the last two power equations:
$$R = \frac{P}{I^2}$$
$$I = \sqrt{\frac{P}{R}}$$
$$E = \sqrt{PR}$$
$$R = \frac{E^2}{P}$$
Notes:Algebra is an extremely important tool in many technical fields. One nice thing about the study of electronics is that it provides a relatively simple context in which fundamental algebraic principles may be learned (or at least illuminated).
The same may be said for calculus concepts as well: basic principles of derivative and integral (with respect to time) may be easily applied to capacitor and inductor circuits, providing students with an accessible context in which these otherwise abstract concepts may be grasped. But calculus is a topic for later worksheet questions . . .
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Question 2 of 13
The Q factor of a series inductive circuit is given by the following equation:
$$Q = \frac {X_L}{R_{series}}$$
Likewise, we know that inductive reactance may be found by the following equation:
$$X_L = 2\pi f L$$
We also know that the resonant frequency of a series LC circuit is given by this equation:
$$f_r = \frac {1}{2 \pi \sqrt{LC}}$$
Through algebraic substitution, write an equation that gives the Q factor of a series resonant LC circuit exclusively in terms of L, C, and R, without reference to reactance (X) or frequency (f).
Reveal answer$$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$$
Notes:This is merely an exercise in algebra. However, knowing how these three component values affects the Q factor of a resonant circuit is a valuable and practical insight!
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Question 3 of 13
We know that the current in a series circuit may be calculated with this formula:
I = Etotal RtotalWe also know that the voltage dropped across any single resistor in a series circuit may be calculated with this formula:
ER = I R Combine these two formulae into one, in such a way that the I variable is eliminated, leaving only ER expressed in terms of Etotal, Rtotal, and R.
Reveal answerER = Etotal ( R Rtotal) Follow-up question: algebraically manipulate this equation to solve for Etotal in terms of all the other variables. In other words, show how you could calculate for the amount of total voltage necessary to produce a specified voltage drop (ER) across a specified resistor (R), given the total circuit resistance (Rtotal).
Notes:Though this “voltage divider formula” may be found in any number of electronics reference books, your students need to understand how to algebraically manipulate the given formulae to arrive at this one.
In a lot of question, the ‘+’ symbol is not printed. Can you please correct this as it causes confusion while trying to solve the equation?
Question 9 says “Combined these two formulae into one solving for the resistance of a copper wire sample (RT) at a specific temperature in degrees Celsius (TC)...”.
I think it should be “at a specific temperature in degrees Fahrenheit (TF)”.