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I am a beginner electronics hobbyist. I found this book called Basic Electronics for Scientists and Engineers and I am currently reading on RC circuits but got stuck trying to understand the equation.

schematic

simulate this circuit – Schematic created using CircuitLab

The given schematic is as shown above and the equation obtained using KVL is

where Q is the charge and C is the Capacitance. (Eq 2.16)

Then it states that

We next take the derivative of Eq. (2.16) in order to remove Q in favor of the current I. Since V is constant for either switch position, we obtain for both cases

This is where I struggled to understand how the second equation is derived from the first equation and my weak calculus knowledge isn't helping either.

I understand that since current is changing and resistance is constant, we get but why did became and why did V became 0?

I would really appreciate it if I could get some clarification or guidance on how to understand this further because I really couldn't figure it out on my own. Thanks in advance.

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    \$\begingroup\$ Differentiating Q/C gives I/C because dQ/dt = I. I feel it may help to review basic calculus and first order ordinary differential equations. \$\endgroup\$
    – across
    Commented Jan 1, 2021 at 7:51
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    \$\begingroup\$ In other words , the definition of current is the rate of change in Q charge flow. \$\endgroup\$ Commented Jan 1, 2021 at 7:55
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    \$\begingroup\$ If this book is spending a lot of time solving KVL and KCL problems, then you should get another book. Try: The Art of Electronics by Paul Horowitz and Winfield Hill. It is more important for a hobbyist to know what happens if you put a square wave into an RC (exponential voltage curve) or a sine wave into an RC (low pass filter). \$\endgroup\$
    – Mattman944
    Commented Jan 1, 2021 at 8:20
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    \$\begingroup\$ I don't know about book but even mathematically this method is not good because 1.Any differential equation depend on input and initial conditions so time at which instant switch is closed or opened and initial condition must be known at that time 2.do not assume V constant rather it should be V u(t) if switch is closed at t=0 \$\endgroup\$
    – user215805
    Commented Jan 1, 2021 at 8:56
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    \$\begingroup\$ @Edley Are you still interested in some discussion of the circuit and calculus? (I think I can make the calculus part pretty easy to gather -- I enjoy teaching it.) Or are you good, now? \$\endgroup\$
    – jonk
    Commented Jan 1, 2021 at 20:40

1 Answer 1

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  1. Since V is assumed constant, the derivative is 0
  2. Q=i*t
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    \$\begingroup\$ V is not constant input ,it is unit step input! and its derivative contain an impulse function \$\endgroup\$
    – user215805
    Commented Jan 1, 2021 at 13:44
  • \$\begingroup\$ Can you please expand your answer? \$\endgroup\$
    – Voltage Spike
    Commented Jan 3, 2021 at 7:26
  • \$\begingroup\$ V is constant. Because of the switch, the initial condition on the capacitor is 0 Volts. Q=I*t, so the derivative with respect to time is I \$\endgroup\$ Commented Jan 4, 2021 at 11:04

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