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I'm solving this question: enter image description here

From this video

We can simplify the circuit to be a single capacitor with capacitance of 50 micro farads.

Using Q = CV, we can determine the charge on each capacitor in series is 4800 micro-coloumbs.

So far so good.

The confusing part:

The 1st capacitor has a charge of 4800 micro coloumbs, as expected

The 2nd "capacitor" is actually a parallel circuit. We calculate the charge of C2 (70 microfarads) using (7/10 * 4800) = 3360 microfarads.

This means the charge of the other path (C3 + C4 in series) is 4800 - 3360 = 1440.

Still, so good.

However, since C3 & C4 are in series, they both must have the same charge. So they both have 1440 micro-coloumbs!

This means the total charge on the circuit composed of (C2,C3,C4) ... adds up to 3360 + (1440 * 2) = 6240, which is greater 4800.

My understanding of what is going is: C3 + C4 should be treated as 1 big capacitor, so the 1440 charge is on both of them at the same time rather than 2 instances of 1440 micro-coulombs on each individual capacitor.

Does this make sense?

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    \$\begingroup\$ If your car was travelling at 50 mph, because you have 4 wheels would you conclude that the total speed is 200 mph? \$\endgroup\$
    – Andy aka
    Nov 2 at 9:16
  • \$\begingroup\$ Answer of GT Electronics is correct. You need to treat parallel capacitors as series resistors and series capacitors as parallel resistors mathematically (sum vs inverted sum of inverses) \$\endgroup\$
    – Ilya
    Nov 2 at 13:47
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Your plan to treat C3 + C4 as "one big capacitor" is wrong. You should treat C3 + C4 as "one SMALL capacitor". A 40uF + 120uF capacitors in SERIES is like one 30uF capacitor. Google "equation for capacitors in Series". Note: This equation is very similar to the equation for Resistors in parallel.

Then, this combination of C3+C4 (30uF) can simply be added to the capacitance of C2 in parallel (70uF) to get your subtotal = 100uF.

The subtotal (100uF) has to be run through the equation for capacitors in Series with C1 (also 100uF) to get the total. Which will be 50uF.

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You somewhat answer you own question at the end there.

The behavior we observed between C1 and the parallel circuit, in which we assign the calculated charge of 4800 uC to each one, is the same behavior we observe between C3 and C4. The charge between two capacitors in series is equal, so your conclusion of looking at it as one instance instead of two might be an easy way of remembering this exact behavior.

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