Adding a Capacitor,or inductor whats change in this circuit?

Thread Starter

DaniKowa

Joined Sep 23, 2020
178
Hi,
i need help to solve a little doubt. In the image below is wired two inductors and two capacitors with different values. So the question is:

The original circuit it consist in one inductor and one capacitor. The signal pass through this component.
I want to add i a second paralleled inductor or capacitor with different value. When the sinusoidal voltage pass this paralleled components is the signal divided to by the number of paralleled component (in this case two so 50%/50%) or it depend by the value of components (in this case 33% in C1/L1 and 66% on C2/L2) ?

I have to understand if adding the second component in parallel how much signal I remove from the first component.I hope my question is clear. Thanks.

Z
 

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Papabravo

Joined Feb 24, 2006
18,424
Capacitors in parallel can be added together to form an equivalent capacitor with a larger value. Example:
1 μF || 2 μF = 1 μF + 2 μF = 3 μF N.B. the double vertical bar '||' means devices connected in parallel

Inductors in parallel split the available current according to the ratio of the inductances. The rule for combining two parallel inductors is to divide the product by the sum. Example:
10 μH || 20 μH = (10 μH * 20 μH) /(10 μH + 20 μH) = 200 (μH)² / (30 (μH)) = 6.67 μH

Note that for inductors, just like for resistors, a parallel combination results in a value that is strictly less than either of the individual components.
 

MrAl

Joined Jun 17, 2014
8,990
Hi,
i need help to solve a little doubt. In the image below is wired two inductors and two capacitors with different values. So the question is:

The original circuit it consist in one inductor and one capacitor. The signal pass through this component.
I want to add i a second paralleled inductor or capacitor with different value. When the sinusoidal voltage pass this paralleled components is the signal divided to by the number of paralleled component (in this case two so 50%/50%) or it depend by the value of components (in this case 33% in C1/L1 and 66% on C2/L2) ?

I have to understand if adding the second component in parallel how much signal I remove from the first component.I hope my question is clear. Thanks.

Z

Hello there,

For capacitances in parallel the total capacitance increases.
For inductances in parallel the total inductance decreases.

However, the behavior of the signal 'passing through' these combinations will be difficult to measure when the output impedance is infinite. That is because the components will appear to pass the entire signal voltage to the load no matter what value they are when the load is infinite.

What you need to do to measure any difference is to add a known load such as a resistive load and also you may have some input impedance that also will make a difference although that is usually small.
See the attached schematic modified with both input and output resistances.

To calculate the output signal it becomes a little different though.
In the case of the capacitors:
Vout=Vin*(Rout*w*C)/sqrt((Rout*w*C+Rin*w*C)^2+1)

and in the case of the inductors:
Vout=Vin*Rout/sqrt(w^2*L^2+(Rout+Rin)^2)

where
C is the total capacitance,
L is the total inductance,
w is 2*pi*f,
f is the frequency in Hertz.

Try some calculations and see what you can find out.
If you post some results here we can then talk about them.

Note that if you have a capacitance meter and inductance meter you may be able to measure the results directly as long as the meters can operate within the range of values you intend to experiment with.
If you measure input and output voltages you can also solve for C and L in those two equations if you want to try that. It's interesting to do it both ways.
 

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Thread Starter

DaniKowa

Joined Sep 23, 2020
178
Thank for responses. I did another simulation and as for the inductors, adding a second inductor a parallel of greater value, the passage of the current is greater in the inductor with the lower value. In my case this is fine because the primary inductor is a input probe where it must have the greatest possible signal input.
 
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