# Proving RC and RL circuits as integrator and differentiator using differential equations [closed]

I need to use first degree ordinary differantial equations to prove (solve) the Differentiator and Integrator Circuits of RL and RC (for example if RL is differeantiator or integaror or both i need to prove it with using first order ordinary differeantial equations).

Btw I know how to use differential equations but i dont know how to use them on circuits so i would be happy if you also show me the solution of some to solve others or maybe hint,tip etc. or some sources where i can learn this about and sorry for my english and this is my first question on this site. Waiting for your answers. :)

## closed as off-topic by uint128_t, ThreePhaseEel, Lorenzo Donati, Peter Smith, pjc50Jan 3 '17 at 15:39

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• Perhaps you could ask what you don't understand about Mr. Wiki's fine math: en.wikipedia.org/wiki/RC_circuit – glen_geek Dec 22 '16 at 16:08
• A simple RC (or RL) is neither a perfect integrator nor a perfect differentiator because of the presence of the resistor. (Hint: don't waste your time trying to prove it). – Andy aka Dec 22 '16 at 16:10
• @Andyaka Well since it'll affect my grades i'll have to. So please dont type things that dont helps me :). – onur cevik Dec 22 '16 at 16:54
• @glen_geek wikipedia doesnt solve it with dif. eq. its just shows it. – onur cevik Dec 22 '16 at 16:55
• I'm voting to close this question as off-topic because this looks like a homework problem with no attempt to solve. Electrical Engineers should be able to attempt to solve problems – Voltage Spike Dec 22 '16 at 18:34

The Photon has shown the basic math relating current to voltage. Look at the differentiator (engineers can't help referring to circuits). At left, the driving function is represented by a Voltage source (Vin), which varies with time: simulate this circuit – Schematic created using CircuitLab

Current is produced by Vin that is common to C1 and to R1. As shown by The Photon, current is proportional to the derivative of Vin. But for this to be true, all of Vin must appear across C1. That leaves no voltage at Vout (voltage across R1 will be zero).
So we compromise a little, and we say that Vout is almost differentiating Vin, so long as Vout is very small. It is an approximation. If Vin has a rate-of-change that is too high, Vout becomes too large, and the approximation fails.
An example Vin:
Let us say that Vin is a ramp whose voltage rises with time...Vin=+10volts/sec. And let us make the assumption that all this voltage appears across C1.
Then I = C * 10, I=10uA
This current will produce a voltage across R of 10uA * 100 ohms = +0.001v. Now if Vin starts at zero volts and rises from there, for awhile it is not larger than 0.001v, so Vout cannot be said to represent a proper differentiated voltage. But later, when Vout rises to a larger voltage (above approximately 0.1v), the approximation becomes more true, and Vout rises toward 0.001v. After awhile, Vin becomes much larger than Vout, and Vout approaches very close to +0.001v.
And if Vin changes too quickly, then Vout will be larger, making the approximation untrue. These circuits only approach their mathematical ideal only when Vout << Vin.

You have an integrator (at right), with Vout representing voltage across the capacitor. Here, R2 must see nearly all of Vin, so that current is equal to Vin/R2. In this case too, Vout << Vin is the required condition for the integration to be true. In this case, Vin must be very much larger than Vout, and/or it must change very quickly.
This integrator fails to integrate a steady voltage at Vin because Vout eventually rises to equal that steady voltage (violating the approximation that Vout << Vin). It really only integrates rapid changes of Vin, so that those changes of Vout are much less than changes of Vin.

Example Vin for the case of Resistor-Inductor
Again, Vin starts at zero and rises linearly at a rate of 10v/sec. For the resistor-inductor differentiator, Vout is taken across the inductor L1(0.01 henries) in the circuit below. And once again, Vout must be very much smaller than Vin. Thus, almost all of Vin appears across R3, so that I = Vin/R3. Now I is rising 0.1A/sec. When this current is applied to inductor L1, a constant voltage is produced...(V = L * di/dt). That voltage is L * I = .01 * .1 = .001 volts.
Notice that this resistor-inductor differentiator yields the same Vout as the resistor-capacitor differentiator above. The time constant of the two circuits are identical...RC = L/R. simulate this circuit

• Thanks for spending time your explanation really helped. I am not good at english but i think i started to understand the point and I'll just try to analyze it more . Btw would it be correct if i just take those integrator and differentiator circuits and apply them kirchoff then combine the answer with dif. eq. on previous comments? And about RL circuits is it just replacing capacitors with inductor or does it have different schematic? – onur cevik Dec 22 '16 at 19:01
• Series RL circuit can differentiate Vin if you take Vout across L, and can integrate if you take Vout across R. As always, Vout << Vin approximation is required. – glen_geek Dec 22 '16 at 19:10
• Thanks again. Looking the schematics on the internet now. Btw i choosed your reply as solution because it covers everything i wondered. I'll try to solve them now. – onur cevik Dec 22 '16 at 19:16
• @onurcevik, whoops, missed my numbers by orders of magnitude, have corrected my answer. (C1 = 1x10^-6) – glen_geek Dec 22 '16 at 21:50
• Thanks for your answers and sorry for late answer.your answer helpmed me the understand it and i used the formulas that The Photon gaved me and kirchoff laws and so far i solved RC as both differentiator and integrator assuming R and C so large for integrator and so low for differentiator. But i stuck with RL i used kirchoff and get equation : Vi=Ldi/dt + Rdq/dt and if i see R as small it gaves me eq: Vi=L(di/dt) and if i integrate it it becomes : 1/L (integral) vi = i which is the point im stuck now. Any idea ? – onur cevik Dec 25 '16 at 9:03

Just as linear resistors are defined by Ohm's Law, there are defining equations for linear capacitors and inductors. For a capacitor

$$Q = CV$$

or, in terms of a derivative

$$I = C\frac{\rm{d}V}{\rm{d}t}$$

For an inductor

$$\frac{\rm{d}I}{\rm{d}t}=\frac{V}{L}$$.

Combine these with Kirchoff's Laws and you will have differential equations describing your circuit.

• +1 for showing that only a pure L or C can integrate or differentiate. – Andy aka Dec 22 '16 at 16:17
• Yeah but how do i solve them to prove RC(for example) as integrator or differentiator? I mean i already started to solve it but i really dont know how to finish it. Our instructor really didnt teached us about it too much all we did was just watching youtube videos on class and he gave us 1 week for this. – onur cevik Dec 22 '16 at 16:58
• @onur, your question says "I know how to use differential equations"... so go ahead from here. Like Andy says, it will only be approximately an integrator or differentiator, not a perfect one. For example, in the RC "integrator" you'll need to consider cases where $v_o \ll v_i$. – The Photon Dec 22 '16 at 17:07
• @ThePhoton The problem is i dont know how to use them on circuits. Thats what my teacher wants so even if will be approximately i have to solve it. So any help is greatly appreciated. And about kirchoffs laws en.wikipedia.org/wiki/Kirchhoff's_circuit_laws – onur cevik Dec 22 '16 at 17:19
• which formula on that link ill use to combine with yours ? – onur cevik Dec 22 '16 at 17:20