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Question

The connections shown below is established with the switch S which is open at present. How much charge will flow through the swiitch if it is closed?


A
15 μC
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B
20 μC
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C
10 μC
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D
30 μC
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Solution

The correct option is A 15 μC
When S is open

C1 and C2 are in series and let their equivalent capacitance is C. Similarly, C3 and C4 are in series and their equivalent capacitance C′′ and C are in parallel.

Let charge flows through C1 and C2 be q1 and charge through C3 and C4 be q2.

q1=CV=3×63+6(10)=20 μC

q2=Cv=3×63+6(10)=20 μC

Thus, q1=q2=q

We can distribute cahrge q1 and q2 in the given circuit as,


When switch S is closed:


From the above fig, we can observe that


By symmetry the potential difference at x will be half of emf given,
x=5 V

Using VL at juction x, shown in fig (ii)

q1+q2+q3=0 ....(1)

The new charge across capacitorsC1 and C2 are,

q1=3(x10)=3(510)=15 μC

q3=6(x0)=6(50)=30 μC

q1 and q2 is the difference of final and initial charge on capacitorC1 and C2 respectiely.

q1=q1(q)=15(20)=5 μC

q3=q3q=3020=10 μC

Substituting the value of q1 and q2in eqn (1), we get

5+q2+10=0

|q2|=15 μC

Hence, option (a) is correct answer.

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