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Question: The connections shown in figure are established with the switch \( S \) open. The charge which flows...

The connections shown in figure are established with the switch SS open. The charge which flows. Through the switch if it is closed is.

(a) 40μC40\mu C
(b) 30μC30\mu C
(c) 20μC20\mu C
(d) 12μC12\mu C

Explanation

Solution

Hint : The flows of charge in an electric circuit signifies. The existence of electric current in that circuit. We can define electric current as the flow of electric charges in a conductor.
When charge flows through the open switch the circuit will not reach or it is an incomplete connection.
When charge is how through the close switch the circuit will react or it is complete connection.
Formula:- We fuse the capacitor formula.
Ceq=(C1+C1C2)1+(C4C3+C4)1{{C}_{eq}}={{\left( {{C}_{1}}+\dfrac{{{C}_{1}}}{{{C}_{2}}} \right)}^{-1}}+{{\left( \dfrac{{{C}_{4}}}{{{C}_{3}}}+{{C}_{4}} \right)}^{-1}}
Q=Ceq.VQ={{C}_{eq}}.V
Where
Q=Q= charge
V=V= voltage
Ceq={{C}_{eq}}= equivalent capacitor .

Complete Step By Step Answer:
Capacitor given
C1=1μF{{C}_{1}}=1\mu F
C2=2μF{{C}_{2}}=2\mu F
C3=2μF{{C}_{3}}=2\mu F
C4=1μF{{C}_{4}}=1\mu F
V=24VV=24V
Ceq=(1+12)1+(12+1)1{{C}_{eq}}={{\left( 1+\dfrac{1}{2} \right)}^{-1}}+{{\left( \dfrac{1}{2}+1 \right)}^{-1}}
=23+23=43μF=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}\mu F
So, Q=Ceq.V=43×24=32μFQ={{C}_{eq}}.V=\dfrac{4}{3}\times 24=32\mu F
This will be divided equally, as the capacitor are equal Q=16μF\Rightarrow Q=16\mu F

(a)

(b)
When the switch SS is closed as shown in the figure (b), then the charges stored across positive plates capacitor 11 and be Q1&Q2{{Q}_{1}}\And {{Q}_{2}} The same charges will be distributed across negative plates of capacitor 33 and 44 Let potential at the negative terminal is zero then the charge across capacitor 1,2,31,2,3 and will be.
Q1=(24V0)×1μF.....(a){{Q}_{1}}=\left( 24-{{V}_{0}} \right)\times 1\mu F.....(a)
Q2=(24V0)×2μF...(b){{Q}_{2}}=\left( 24-{{V}_{0}} \right)\times 2\mu F...(b)
Q3=V0×1μF...(c){{Q}_{3}}={{V}_{0}}\times 1\mu F...(c)
Q4=V0×1μF...(d){{Q}_{4}}={{V}_{0}}\times 1\mu F...(d)
Thus from (a)(a) and (d)(d) V0=12V{{V}_{0}}=12V and thus the charges be: Q1=Q4=12μF&Q2=Q3=24μF{{Q}_{1}}={{Q}_{4}}=12\mu F\And {{Q}_{2}}={{Q}_{3}}=24\mu F
In conclusion,
Thus when the switch is zero, 12μF12\mu F charge is passed through the switch.

Additional Information:
In this question we used the capacitor formula and found the state at which switch is closed and with value of charges which will flow through the closed switch by using formula.

Note :
Capacitance is the ratio of the amount of electric charge stored on a conductor to a difference in electric potential. There are two closely related motions of capacitance: self-capacitance and mutual capacitance.