Switch S, shown in the figure below, is closed after having been open for a long
ID: 1770317 • Letter: S
Question
Switch S, shown in the figure below, is closed after having been open for a long time.
1)
What is the initial value of the battery current just after switch S is closed?
A
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2)
What is the battery current a long time after switch S is closed?
A
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3)
What are the charges on the plates of the capacitors a long time after switch S is closed?
Q5µF=µC
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4)
Q10µF =C
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5)
Switch S is reopened. What are the charges on the plates of the capacitors a long time after switch S is reopened?
Q5µF=µC
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6)
Q10µF =µC
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Explanation / Answer
1) just after the switch is closed , the capacitors will be short circuited
hence , for total resistance
Rnet = 10 + 1/(1/15 + 1/15 + 1/12)
Rnet = 14.62 Ohm
initial current in the battery = 50/Rnet
initial current in the battery = 50/14.62
initial current in the battery = 3.42 A
2)
after a very long time ,
the capacitors will be open circuited
Rnet = 10 + 15 + 15 + 12 = 52 Ohm
current in battery after long time = 50/52
current in battery after long time = 0.96 A
3)
for the 5 uF capacitor
charge on 5 uF capacitor = 5 * (15 + 12) * 0.96
charge on 5 uF capacitor = 129.6 uC
4)
for the 10 uF capacitor
charge on 10 uF capacitor = 10 * (15 + 12) * 0.96
charge on 10 uF capacitor = 259 uC
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