A 3.00 - µF and a 4.00 - µF capacitor are connected in series, and this combination is connected in parallel with a 2.00 - µF capacitor (see the figure ). What is the net capacitance?
3.00 - µF 4.00 - µF
2.00 - µF
26.0 V
For the two caps in series, 1/C = 1/3 + 1/4
C = 12/7
For the two caps now in parallel, C = 2 + 12/7 = 26/7 = 3.71
Thank you Steve!
2.0
4th
Well, in this case, the net capacitance is like trying to calculate the total enthusiasm at a circus with multiple clowns. You see, when capacitors are connected in series, it's like having a group of serious clowns who are trying to balance on a tiny little unicycle together. They may be skilled individually, but together, it's quite a challenging act!
To calculate the total capacitance in a series combination, you need to use the formula:
1/C = 1/C1 + 1/C2 + 1/C3
In our case, C1 is 3.00µF, C2 is 4.00µF, and C3 is 2.00µF. So, let's plug in these values and calculate!
1/Total_Capacitance = 1/3.00µF + 1/4.00µF + 1/2.00µF
After calculating, you'll find that the total capacitance is approximately 1.11µF. So, it's like ending up with a clown car filled with enthusiastic clowns who managed to balance themselves on that tiny unicycle. Impressive, right?
To find the net capacitance of the given capacitors, you need to understand how capacitors behave when connected in series and parallel.
When capacitors are connected in series, the total capacitance (Cs) is given by the reciprocal of the sum of the reciprocals of individual capacitances:
Cs = 1 / (1/C1 + 1/C2)
In this case, you have a 3.00 µF and a 4.00 µF capacitor in series:
Cs = 1 / (1/3.00 + 1/4.00)
= 1 / (0.3333 + 0.25)
= 1 / 0.5833
≈ 1.714 µF
So the combined capacitance of the 3.00 µF and 4.00 µF capacitors in series is approximately 1.714 µF.
Now, the combined capacitance of the capacitors in series is connected in parallel with a 2.00 µF capacitor. When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances:
Cp = 1/ (1/Cs + 1/C3)
Where Cp is the total capacitance and C3 is the capacitance of the third capacitor.
In this case, the series combination of 3.00 µF and 4.00 µF capacitors has a total capacitance of 1.714 µF, and it is connected in parallel with a 2.00 µF capacitor:
Cp = 1 / (1/1.714 + 1/2.00)
= 1 / (0.5828 + 0.5)
= 1 / 1.0828
≈ 0.923 µF
Therefore, the net capacitance of the given combination of capacitors is approximately 0.923 µF.