In the figure below( Figure 1 ) , each capacitor has C = 4.50 ? and V ab=23.5V.
ID: 1287780 • Letter: I
Question
In the figure below(Figure 1) , each capacitor has C = 4.50? and Vab=23.5V.
a) Calculate the charge on each capacitor. Enter your answers in ascending order separated by commas.
b) Calculate the potential difference across each capacitor.
Enter your answers in ascending order separated by commas.
In the figure below(Figure 1) , each capacitor has C = 4.50? and Vab=23.5V. a) Calculate the charge on each capacitor. Enter your answers in ascending order separated by commas. b) Calculate the potential difference across each capacitor. Enter your answers in ascending order separated by commas.Explanation / Answer
The top two C's are in series
1/C = 1/C1 + 1/C2
1/C = 1/4.5 + 1/4.5
C = 2.25 F
That is in parallel with another 4.5, so C = 4.5 + 2.25 = 6.75
That is in series with the final C
1/C = 1/6.75 + 1/4.5
Total C = 2.7 F
Now we can find the charge per capacitor
Apply Q = CV
Q = (2.7)(23.5)
Q = 63.45 C (This is the charge on the bottom C)
The voltage on that C = 63.45/4.5 = 14.1 V
The voltage parallel portion
V = 63.45/6.75 = 9.4 V (This is the V on the third Capacitor)
For this one Q = CV = (4.5)(9.4) = 42.3 C
The Q on the upper two...
Q = 9.4(2.25) = 21.15 C
Then the Voltage on each of those is V = Q/C = 21.15/4.5 = 4.7 V
So, in summary...
Part A) 21.15 C, 21.15 C, 42.3 C, and 63.45 C
Part B) 4.7 V, 4.7 V, 9.4 V & 14.1 V
This solution is based on the Capacitors of 4.5 Farads. I asked about what the question mark means. If it is not Farads, comment below and I can make the correction. The units will make a difference.