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I know it may sound stupid but having thinking across my mind with the basic knowledge I've already have with capacitance(series and parallel), I've been wondering on how to get the total capacitance in a circuit with other electronic components as shown below.

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where C1 - 2500uF, C2 - 50uF, C3 - 20uF, C4 - 240uf, C5 - 50nF

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    \$\begingroup\$ 1) How do you mean "total capacitance"? Capacitance is only defined between 2 nets, which nets do you mean, the output nets? 2) Why would you need to know that value? When you know that value, how/what will that help? \$\endgroup\$ Dec 5, 2019 at 13:10
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    \$\begingroup\$ "I've been wondering on how to get the total capacitance in a circuit " - add all the capacitors is the only (but pointless) answer I can think of. \$\endgroup\$
    – Andy aka
    Dec 5, 2019 at 13:16
  • \$\begingroup\$ Or are you talking about effective capacitance if this is considered as capacitance multiplier circuit? \$\endgroup\$
    – Justme
    Dec 5, 2019 at 14:44

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When capacitors are not wired directly together the 'total capacitance' of the circuit depends on what other components are between them. The other components could add resistance or inductance, and then the impedance at the port of interest will not be purely capacitive.

In your example circuit, if voltage adjustment pot R2 is wound fully down then C3 is shorted out so it doesn't contribute any capacitance to the circuit. If the pot is wound fully up then C2 is directly in parallel with C3, giving a combined capacitance of 50uF + 20uF = 70uF. However C2 is also connected in parallel with C1 via R1, which produces a combined capacitance of 2500uF + 70uF = 2570uF.

Now consider the capacitors on the other side of Q1. Under normal operating conditions the Collector to Emitter impedance is high, but not infinite. The Base to Emitter junction also has some impedance, so C4 and C5 are also (weakly) in parallel with C1 and C2 etc. through Q1, giving a 'total capacitance' of 2570uF + 240uF + 50nF = 2810.05uF (this is not the full story because Q1 also 'amplifies' the capacitance on the Base at the Emitter, but for simplicity we will ignore this effect).

So the 'total capacitance' seen by the rectifier is 2810.05uF, but the impedance is not totally capacitive because there are resistances between and across some of the capacitors. Therefore C1 will have the dominant effect because it is much larger than the others and does not have any resistance between it and the rectifier. On the other hand at the output side C4 has the dominant effect.

For basic understanding of a circuit you can generally consider capacitors that are not connected through small resistances to be separate from each other, though they may still interact slightly. In this circuit the values of C1, C2, C3 and C4 would be chosen according to their local effect, because calculating the total impedance at all points in the circuit is tedious and unnecessary. To see what effect they have without doing a lot of math you could put the circuit into a simulator such as LTspice, then change the values and see what happens.

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  • \$\begingroup\$ Just another question, why do we put some capacitors along the base collector side, when the effect could be only seen at the output side? \$\endgroup\$
    – merP
    Dec 6, 2019 at 0:00

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