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schematic

simulate this circuit – Schematic created using CircuitLab

Question is, how to calculate capacitance of such a circuit. I made a schematic with resistors (because that i can measure) and from there i saw that the C4 value is always ignored, but i can't be sure the capacitors will act the same. I have been trying to break it down into smaller circuits, but the C4 just makes me hold my head.

How should i break it down into series or parallel capacitors so i can actually calculate the total capacitance?

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    \$\begingroup\$ your schematic is badly drawn .... line up C2 and C5 horizontally ... line up C3 and C6 horizontally ... keep C4 in midde ... turn C4 quarter turn (90°) \$\endgroup\$ – jsotola Dec 22 '17 at 4:45
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    \$\begingroup\$ how can i put them horizontally when there is a wire inbetween \$\endgroup\$ – s3v3ns Dec 22 '17 at 4:56
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    \$\begingroup\$ redrew the schematic ... waiting for review \$\endgroup\$ – jsotola Dec 22 '17 at 19:04
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    \$\begingroup\$ The first thing you should ask yourself is this: Does \$C_4\$ matter? Or can you defuse it? - The total capacitance of this circuit is 4 µF, just saying. \$\endgroup\$ – Harry Svensson Dec 22 '17 at 21:42
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Quick observation -- This is a balanced Wheatstone bridge circuit of capacitors. Because,

$$C_2/C_3 = C_5/C_6$$

So no charge will across C4. You can just neglect it. So the effective capacitance would be simply :-

$$C_{eq} = (C_2||C_5)+(C_3||C_6)$$

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enter image description here The capacitor C4 does not appear in the equivalent capacitance of the circuit. This is because the circuit is symmetrical around the two ends of the capacitor and there is no reason for the voltage to be different across these points. This would imply a zero potential difference (and charge) across the capacitor which can thus be removed without affecting the overall circuit.
But you can still apply KVL and charge equations to arrive at the same result:
Suppose the battery transfers a charge Q into the capacitors and the charge distribution across the capacitors looks as in the figure above: Applying KVL across C2-C5-C6-C3 loop, we get: $$\frac{q1}{C} + \frac{q2}{C} = \frac{Q-q1}{C} + \frac{Q-q2}{C}$$ $$\implies q1 + q2 = Q$$ Another KVL across C2-C1-C3 loop gives: $$-\frac{q1}{C} - \frac{q1-q2}{C1} + \frac{Q-q1}{C} = 0$$ $$\implies (\frac{1}{C1} + \frac{1}{C})(q1-q2) = 0$$ Solving these equations we get, $$q1 = q2$$ Thus, there is no charge across the C6 capacitor concluded before.
It is now pretty straight forward to solve for the equivalent capacitance now.

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It is quiet easy to prove whether C4 has any effect on total capacitance when C2=C3=C5=C6. Here's your circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I will calculate total capacitance Ct of this circuit and if the final result does not contain Cx than it can be omitted, it has no effect. But if the final result contains Cx that it has some effect and cannot be omitted.

First step is to use \$ \Delta - Y \$ transformation:

schematic

simulate this circuit

\$C1=(Ca\cdot Cb+Cb\cdot Cc+Ca\cdot Cc)/Ca\$
\$C2=(Ca\cdot Cb+Cb\cdot Cc+Ca\cdot Cc)/Cb\$
\$C3=(Ca\cdot Cb+Cb\cdot Cc+Ca\cdot Cc)/Cc\$

So after transformation our circuit will look like this:

schematic

simulate this circuit

where
\$C1=(Cx\cdot C+C\cdot C+Cx\cdot C)/Cx=(2\cdot C\cdot Cx+C\cdot C)/Cx\$ \$C2=(Cx\cdot C+C\cdot C+Cx\cdot C)/C=2\cdot Cx+C\$ \$C3=(Cx\cdot C+C\cdot C+Cx\cdot C)/C=2\cdot Cx+C\$

After this transformation we can easily calculate total capacitance of this circuit.
Capacitance of C3 and C in series:
\$C3c=\frac{C3\cdot C}{C3+C}=\frac{C(2\cdot Cx+C)}{2(Cx+C)}\$
Capacitance of C2 and C in series:
\$C2c=\frac{C2\cdot C}{C2+C}=\frac{C(2\cdot Cx+C)}{2(Cx+C)}\$

schematic

simulate this circuit

C3c and C2c are in parallel so it is easy to join:
\$C32=C3c+C2c=\frac{C(2\cdot Cx+C)}{2(Cx+C)}+\frac{C(2\cdot Cx+C)}{2(Cx+C)}=\frac{C(2\cdot Cx+C)}{(Cx+C)}\$

And now the grand finale, the total capacitance (C1 and C32 in series):
\$Ct=\frac{C1\cdot C32}{C1+C32}=\frac{\frac{2\cdot C\cdot Cx+C\cdot C}{Cx}\cdot \frac{C(2\cdot Cx+C)}{(Cx+C)}}{\frac{2\cdot C\cdot Cx+C\cdot C}{Cx}+\frac{C(2\cdot Cx+C)}{(Cx+C)}}=\frac{\frac{C^2(2\cdot Cx+C)^2}{Cx(Cx+C)}}{\frac{C(2\cdot Cx+C)^2}{Cx(Cx+C)}}=C\$

So, as we can see, the capacitance (Ct) of the circuit is represented only by value of C, and there is no Cx, so it is proved that Cx has no effect on that circuit at all and can be omitted.

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