Question

Determine the equivalent capacitance between a and b for the group of capacitors in the drawing.

Answer 1

Answer:

There are different of arrangements for the capacitors to work inside a given system, as there are two different form of arrangements for the capacitors to be in, which are:

1. The parallel connections of capacitor,
2. The series connections of capacitor.

Explanation:

The parallel connection of capacitor is comprised of capacitors having the unequal amount of current, ΔI. And has a same potential difference across the number of capacitors. And the equivalent amount of capacitance,ΔC is calculated by the following equation:

• ΔC(eq)=C₁ + C₂ + C₃ + C₄ + C(n),

For the series form of arrangements of the capacitors we have the different form of equation which can be explained by the total number of capacitors that are connected in series in combination as they are provided by the equal amount of current,ΔI across it, while the voltage or total amount of potential difference, Δv provided to the system is more different for each of the capacitor. And the equivalent amount of capacitance, ΔC(equivalent) can be determined by the following equation:

• 1/C(eq)=1/C₁+1/C₂+1/C₃+1/C(n).

Answer 2

Answer:

The total effective capacitance is 2.033 μF

Explanation:

As the complete question is not given, the complete question is found online and is as attached herewith.

As with resistors start as far away from where you want total capacitance to be, so we have 3 capacitors in series, the 24 μF, the 12 μF, and the 8 μF

$\dfrac{1}{C_{effective1}}=\dfrac{1}{C_{24}}+\dfrac{1}{C_{12}}+\dfrac{1}{C_{8}}=\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{8}=\dfrac{1}{4}\\C_{effective1}=4 \mu F$

Next we have two 4 μF capacitors in parallel so we can combine them to

$C_{effective2}=C_4+C_{effective1}\\C_{effective2}=4+4\\C_{effective2}=8 \mu F$

Now we just have three capacitors in series so the total is

$\dfrac{1}{C_{effective1}}=\dfrac{1}{C_{5}}+\dfrac{1}{C_{effective2}}+\dfrac{1}{C_{6}}=\dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{6}=0.49166\\C_{effectivet}=1/0.49166=2.033 \mu F$

So the total effective capacitance is 2.033 μF