Switch S, shown in the figure below, is closed after having been open for a long
ID: 1872326 • Letter: S
Question
Switch S, shown in the figure below, is closed after having been open for a long time.
(a) What is the initial value of the battery current just after switch S is closed?
A
(b) What is the battery current a long time after switch S is closed?
A
(c) What are the charges on the plates of the capacitors a long time after switch S is closed?
(d) Switch is reopened. What are the charges on the plates of the capacitors a long time after switch S is reopened?
Explanation / Answer
a) as the swtich is closed , the capacitors will acts as short circuit
hence ,
current in battery = 50/(10 + 1/(1/15 + 1/15 + 1/12))
current in battery= 3.42 A
b)
after a long time
capacitor will be open circuited
Current in battery = 50/(15 + 12 + 15 + 10)
Current in battery = 0.96 A
c)
for the charges on the capacitor ,
Q(5 uF) = 5 * (12 +15) * 0.96 uC
Q(5 uF) = 130 uC
Q(10 uF) = 10 * (12 +15) * 0.96 uC
Q(10 uF) = 260uC
d) after a long timne after reopening
both capacitors will be fully discharged
charge on both capcitors will be zero
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