Answer to Question #310035 in Physics for kelly

Question #310035
  1. Three capacitors with individual capacitances of 2 μF, 5 μF, and 10 μ respectively are connected in series with a 12V battery. What are the total capacitance and total charge in the network?
  2. Same capacitors in number 1 connected in a parallel. If the combined charge in the network is 50 μC, what are the total voltage and total capacitance in the network?
1
Expert's answer
2022-03-13T18:48:47-0400

1. The total capacitance CC of series connection is given as follows:


1C=12×106F+15×106F+110×106FC=1.25×106F\dfrac{1}{C} = \dfrac{1}{2\times 10^{-6}F} + \dfrac{1}{5\times 10^{-6}F} + \dfrac{1}{10\times 10^{-6}F}\\ C = 1.25\times 10^{-6}F

The total charge:


Q=CV=1.25×106F12V=15×106CQ = CV = 1.25\times 10^{-6}F\cdot 12V = 15\times 10^{-6}C

2. The total capacitance of parallel connection:


C=2×106F+5×106F+10×106F=17×106FC = 2\times 10^{-6}F+5\times 10^{-6}F+10\times 10^{-6}F = 17\times 10^{-6}F

The voltage:


V=QC=50×106C17×106F2.9VV = \dfrac{Q}{C} = \dfrac{50\times 10^{-6}C}{17\times 10^{-6}F} \approx 2.9V

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