Three capacitors of capacitance C1=2.00 μF, C2 =8.00 μF, and C3=10.0 μF are connected to a 40.0 V battery as shown in the figure.
1. Calculate the charge on C3.
2. Calculate the voltage across C1.
Really stuck here - I've tried calculating this multiple ways and still keep doing something wrong. Any help would be appreciated!
Three capacitors at capacitance?
回答 (3)
2015-07-12 2:41 am
✔ 最佳答案
C1 and C2 are in parallel. Parallel capacitors add like series resistors, C = 2 uF + 8 uF = 10 uFThese are in series with C3. Series capacitors add like parallel resistors. 1/C = (1/10) + (1/10)
1/C = 2/10
C = 5 uF
So the circuit is equivalent to a single capacitor of 5 uF. Use C = Q/V or Q = CV to calculate the charge Q which appears on this equivalent capacitor.
That charge appears on the side of C3 closest to the battery, and therefore the other side of C3 as well. So that's the answer to question 1.
You can use Q = CV with C = C3 to find the voltage V3 across C3. The answer to question 2 is the difference between V and V3.
2015-07-12 9:10 am
C12 = C1+C2 = 2.00+8.00 = 10.00 μF
C123 = C12//C3 = (10*10)/(10+10) = 5.00 μF
V1 = V2 = V*C3/(C3+C12) = 40*10/(20) = 20.00 V
V3 = 40-20.00 = 20.00 V
Q3 = V3*C3 = 20.00*10*10^-6 = 200.00 μCoulomb
C123 = C12//C3 = (10*10)/(10+10) = 5.00 μF
V1 = V2 = V*C3/(C3+C12) = 40*10/(20) = 20.00 V
V3 = 40-20.00 = 20.00 V
Q3 = V3*C3 = 20.00*10*10^-6 = 200.00 μCoulomb
2015-07-12 2:29 am
idk
收錄日期: 2021-05-02 11:17:42
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https://hk.answers.yahoo.com/question/index?qid=20150711182827AAU3eWC