All About Circuits

Mathematics for Electronics

Calculus for Electric Circuits


30 questions By Tony R. Kuphaldt

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  • Question 10 of 30


    ∫f(x) dx Calculus alert!

    Ohm’s Law tells us that the amount of current through a fixed resistance may be calculated as such:


    I = E

    R

    We could also express this relationship in terms of conductance rather than resistance, knowing that G = 1/R:


    I = EG

    However, the relationship between current and voltage for a fixed capacitance is quite different. The “Ohm’s Law” formula for a capacitor is as such:


    i = C de

    dt

    What significance is there in the use of lower-case variables for current (i) and voltage (e)? Also, what does the expression [de/dt] mean? Note: in case you think that the d’s are variables, and should cancel out in this fraction, think again: this is no ordinary quotient! The d letters represent a calculus concept known as a differential, and a quotient of two d terms is called a derivative.

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  • Question 11 of 30


    ∫f(x) dx Calculus alert!

    Capacitors store energy in the form of an electric field. We may calculate the energy stored in a capacitance by integrating the product of capacitor voltage and capacitor current (P = IV) over time, since we know that power is the rate at which work (W) is done, and the amount of work done to a capacitor taking it from zero voltage to some non-zero amount of voltage constitutes energy stored (U):


    P = dW

    dt


    dW = P dt


    U = W = ⌠ ⌡ P dt

    Find a way to substitute capacitance (C) and voltage (V) into the integrand so you may integrate to find an equation describing the amount of energy stored in a capacitor for any given capacitance and voltage values.

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  • Question 12 of 30


    ∫f(x) dx Calculus alert!

    Integrator circuits may be understood in terms of their response to DC input signals: if an integrator receives a steady, unchanging DC input voltage signal, it will output a voltage that changes with a steady rate over time. The rate of the changing output voltage is directly proportional to the magnitude of the input voltage:



    A symbolic way of expressing this input/output relationship is by using the concept of the derivative in calculus (a rate of change of one variable compared to another). For an integrator circuit, the rate of output voltage change over time is proportional to the input voltage:


    dVout

    dt
    ∝ Vin

    A more sophisticated way of saying this is, “The time-derivative of output voltage is proportional to the input voltage in an integrator circuit.” However, in calculus there is a special symbol used to express this same relationship in reverse terms: expressing the output voltage as a function of the input. For an integrator circuit, this special symbol is called the integration symbol, and it looks like an elongated letter “S”:


    Vout ⌠ ⌡ T 0 Vin dt

    Here, we would say that output voltage is proportional to the time-integral of the input voltage, accumulated over a period of time from time=0 to some point in time we call T.

    “This is all very interesting,” you say, “but what does this have to do with anything in real life?” Well, there are actually a great deal of applications where physical quantities are related to each other by time-derivatives and time-integrals. Take this water tank, for example:



    One of these variables (either height H or flow F, I’m not saying yet!) is the time-integral of the other, just as Vout is the time-integral of Vin in an integrator circuit. What this means is that we could electrically measure one of these two variables in the water tank system (either height or flow) so that it becomes represented as a voltage, then send that voltage signal to an integrator and have the output of the integrator derive the other variable in the system without having to measure it!

    Your task is to determine which variable in the water tank scenario would have to be measured so we could electronically predict the other variable using an integrator circuit.

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