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Algebraic Equation Manipulation for Electric Circuits
19 questions By Tony R. Kuphaldt
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Question 4 of 19
The following equations solve for the output voltage of various switching converter circuits (unloaded), given the switch duty cycle D and the input voltage:
Vout = D Vin (Buck converter circuit) Vout = Vin 1 − D(Boost converter circuit) Vout = D Vin 1 − D(Inverting or Cuk converter circuit) Manipulate each of these equations to solve for duty cycle (D) in terms of the input voltage (Vin) and desired output voltage (Vout). Remember that duty cycle is always a quantity between 0 and 1, inclusive.
Reveal answerD = Vout Vin(Buck converter circuit) D = 1 − ( Vin Vout) (Boost converter circuit) D = Vout Vin + Vout(Inverting or Cuk converter circuit) Notes:Given the equations for these converter circuit types solving for output voltage in terms of input voltage and duty cycle D, this question is nothing more than an exercise in algebraic manipulation.
Note to your students that all of these equations assume a condition of zero load on the converter circuit. When loads are present, of course, the output voltage will not be the same as what is predicted by these neat, simple formulae. Although these DC-DC power converter circuits are commonly referred to as “regulators,” it is somewhat misleading to do so because it falsely implies a capacity for self-correction of output voltage. Only when coupled to a feedback control network are any of these converter circuits capable of actually regulating output voltage to a set value.
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Question 5 of 19
Solve for n in the following equations:
Equation 1: −56 = −14n
Equation 2: 54 − n = 10
Equation 3: 4/n = 12
Equation 4: 28 = 2 − n
Reveal answerEquation 1: n = 4
Equation 2: n = 44
Equation 3: n = 0.[333]
Equation 4: n = −26
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 a in each equation. Remind them to document each and every step in the process, so that nothing is left to guess or to chance.
Equations 2 through 4 require two steps to solve for n. Equation 1 only requires a single step, but the two negative numbers may be a bit confusing to some.
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Question 6 of 19
The formula for calculating total resistance of three series-connected resistors is as follows:
$$R = R_1+R_2+R_3$$
Algebraically manipulate this equation to solve for one of the series resistances (R1) in terms of the other two series resistances (R2 and R3) and the total resistance (R). In other words, write a formula that solves for R1 in terms of all the other variables.
Reveal answer$$R_1 = R - (R_2+R_3) \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ R_1 = R - R_2 - R_3$$
Notes:This question is nothing more than practice algebraically manipulating equations. Ask your students to show you how they solved it, and how the two given answers are equivalent.
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.