Switch S, shown in the figure below, is closed after having been open for a long

Question

Switch S, shown in the figure below, is closed after having been open for a long time.

Q5 Switch S, shown in the figure below, is closed after having been open for a long time. What is the initial value of the battery current just after switch S is closed? A What is the battery current a long time after switch S is closed? A What are the charges on the plates of the capacitors a long time after switch S is closed? Switch is reopened. What are the charges on the plates of the capacitors a long time after switch S is reopened? I give priority to first correctly answered question. I can't give everyone the five stars for effort that I wish I could. Good luck!

Explanation / Answer

a) initially the two capacitors are short circuited

Rnet = 10 + 1/(1/15 +1/12+1/15)) = 14.615 ohms

I = V/Rnet = 50/14.615 = 3.41 A


b)after a long time the two capacitors are opened

Rnet = 10 + 15 + 12 + 15 = 52 ohms

I = V/Rnet = 50/52 = 0.9615 A

c)
potentai difference across two capacitors, = 0.9615*(12+15) = 25.96 volts
Q(5micro F) = 5*10^-6*25.96 = 129.8 micro C
Q(10micro F) = 10*10^-6*25.96 = 259.6 micro C

d)

after a long time charge on each capacitor = (129.8+259.6)/2 = 194.6 micro C

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