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A capacitor C1=1μF is charged up to a voltage V=60V by connecting it to a battery B through the switch (1). Now C1 is disconnected from battery and connected to a circuit consisting of two uncharged capacitors C2=3μF and C3=6μF through a switch (2) as shown in the figure. The sum of final charges an C2 and C3 is:
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(A) 40μC
(B) 36μC
(C) 20μC
(D) 54μC

Answer
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Hint: The sum of the final charges an C2 and C3 can be determined by using two formulas. The capacitor is series formula and then the potential difference in capacitor formula. By using the given information in these two formulas, the sum of final charges C2 and C3 can be determined.

Formulae Used:
Capacitor in series,
1C=1C1+1C2
Where, C is the equivalent capacitance, C1 capacitance of the first capacitor and C2 is the capacitance of the second capacitor.
Potential difference in capacitor,
V=QC
Where, V is the potential difference, Q is the charge and C is the capacitance.

 Complete step-by-step solution:
Given that,
A capacitor, C1=1μF
Voltage V=60V
The capacitance is, C2=3μF and C3=6μF
Capacitor in series, and the capacitors C2 and C3 are in series then,
1C=1C2+1C3..................(1)
By rearranging the terms, then the above equation is written as,
1C=C3+C2C3×C2
By taking reciprocal on both sides of the above equation, then the above equation is written as,
C=C3×C2C3+C2
By substituting the terms given terms in the above equation, then
C=6×36+3
On further simplification, then
C=189
On dividing, then the above equation is written as,
C=2μF
Now, the potential difference across three capacitors C1 and C (Here C is the combined capacitance of C2 and C3 )
V=QC1+C
By substituting the charge and the capacitance in the above equation, then
V=601+2
On further simplification, then
V=603
On dividing the terms in the above equation, then
V=20V
The sum of the charge of the capacitors C2 and C3 are,
Q=C×V (Here C is the combined capacitance of C2 and C3 )
Now, substituting the combined capacitance of C2 and C3, and the potential difference V in the above equation, then
Q=2×20
On multiplying, then
Q=40μC
Thus, the above equation shows the sum of final charges C2 and C3.
Hence, the option (A) is correct.


Note:- The capacitors C2 and C3 are in series, so the capacitor in series formula is used. Then, we have to find the total charge across the second capacitor and the third capacitor, so the equivalent capacitance of C2 and C3 are multiplied with the potential difference.
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