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Quiz 1 - Purely Resistive Circuit


For this introduction to alternating current (AC) circuit's analysis quiz, students are to mathematically solve the following AC series circuit problem. Hint: the answer will be in the form of a math formula.

1. Given a series circuit with an inductor, capacitor, and resistor; determine at what frequency will the circuit appear purely resistive?


Solution:

Because it was stated the circuit would be completely resistive, the impedance for the complex components (inductor and capacitor) would be equal to zero, and therefore could be written computationally as:


XC = 0 and XL = 0

Therefore, XC = XL

With XC = 1 / omega C, where omega = 2pi (frequency)

XC = 1 / 2pi (frequency) C

And,

XL = omega L = 2pi (frequency) L

Setting these two equations equal to one another we would have:

1 / 2pi (frequency) C = 2pi (frequency) L

Now solving for the frequency by isolating it on one side of the equation,

1 = [2pi (frequency) C] [2pi (frequency) L]

1 = (frequency)2[2pi C] [2pi L]

(frequency)2 = 1 / [2pi C] [2pi L]



We would then take the square root of both sides of the equation and get,

frequency = 1 / ( [2pi C] [2pi L] )1/2

Therefore, the series circuit given one inductor, one capacitor, and one resistor (or any combination that can be resolved into an equivalent circuit with one total inductance, total capacitance, and one total impedance) would appear purely resistive when the frequency is equal to the inverse of two times pi, times the square-root of the inverse of the total capacitance times the total inductance.

Do Note, that there is no resistive component (or R value) in the calculation. Why? This is because we are looking for what values force the capacitance and inductive components in the circuit to zero, leaving just the resistive component in the circuit, and this would occur at specific frequencies based on the fixed values of the total capacitance and total inductance in the circuit.




For example, lets say the capacitance is equal to 22uF (or 22 micro Farads), and the inductance is equal to 4.3uH (or 4.3 micro Henrys). Based on the calculation we performed above we would have the following:

frequency = 1 / ( [2pi (2.2uF)] [2pi (4.3uH)] )1/2

Solving for the frequency we would find that,

frequency = 51.75kHz

So what does mean? Ok, if your input AC source was at a frequency of 51.75 kilohertz, the circuit would appear to have no capacitor or inductor values in the circuit; mathematically this is expressed as the imaginary component (remember the complex math expression of 0.0 + j9.9, we’re talking about the + j9.9 value here or in polar form the angular component).

How is this helpful? Well lets say your AC source is 60Hz which is a typical value in the real world. We can do a little math and solve the original expression for the inductor value which would be:

L = 1 / C (2 pi frequency)2

If we say the circuit capacitance in this example is 1000uF, and we defined the frequency to be 60Hz, the inductance to have a purely resistive circuit is:

L = 1 / (1000uF) (2 pi (60Hz))2

L = 7.04mH (the inductance needed to balance out the capacitive component of the circuit)

You will discover in advanced courses how important this simple concept becomes when trying to balance a circuit for a specific purpose. This basic concept is heavily used in electrical power transmission lines and all the way down to the integrated circuits level of design.



Engineering key words: circuits quiz, solutions, answers, alternating current, resistor network, Thevenin, Norton, equivalent circuit, polar, resistance, inductance, capacitance, RL, RC, RLC, passive component, total impedance, cap, delta, calculation, admittance and susceptance, phase and complex notation, power, watts, voltage, current, peek to peek, frequency, load, max and min.


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