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Question
Each capacitor shown in figure has a capacitance of 5⋅0 µF. The emf of the battery is 50 V. How much charge will flow through AB if the switch S is closed?
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Solution
Initially, when the switch S is open, the equivalent capacitance is given by
`C_(eq) = (2CxxC)/(3C)`
`⇒ C_("eq") = 2/3 C = 2/3 xx 5.0 "uF"`
The Charge supplied by the battery is given by
`Q = C_(eq) xx V`
`⇒ Q = 2/3 xx (5.0 "uF") xx (50 "V")`
`⇒Q = 500/3 "uC"`
When the switch S is closed, no charge goes to the capacitor connected in parallel with the switch.
Thus, the equivalent capacitance is given by `C_(eq) = 2C = 2 xx 5.0 xx 10 = uF`
The charge supplied by the battery is given by
`Q = 10 "uF" xx 50 = 500 "uC"`
The initial charge stored in the shorted capacitor starts discharging."?
Hence, the charge that flows from A to B is given by
`Q_"net" = 500 "uC" - 500/3 "uC"`
`⇒ Q_"net" = 3.3 xx 10^-4 C`
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