A capacitor of capacitance 100 - F is connected across a battery of emf 6-0 V through a resistance of 20 k - for 4-0 s- The battery is then replaced by a thick wire- What will be the charge on the capacitor 4-0 s after the battery is disconnected -

C = 100 μ F, E μ f = 6V, r = 20 K, T = 4 s Charging, Q = CV ( 1 e^ t/RC ) = 6 × 10^ 4 (1 e^ 2) = 5.187 × 10^ 4 C Discharging Q = q ( e^ t/RC ) = 5.184 × 1 ...

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Standard XII

Physics

Charging of Capacitors

A capacitor o...

Question

A capacitor of capacitance 100 $μ$ F is connected across a battery of emf 6.0 V through a resistance of 20 k $Ω$ for 4.0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0 s after the battery is disconnected ?

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Solution

$c = 100 μ F, E μ f = 6 V,$

$r = 20 K, T = 4 s$

$Charging,$

$Q = C V (1 - e^{- \frac{t}{R C}})$

$= 6 \times 10^{- 4} (1 - e^{- 2})$

$= 5.187 \times 10^{- 4} C$

$Discharging$

$Q = q (e^{-} \frac{t}{R C})$

$= 5.184 \times 10^{- 4} \times e^{- 2}$

$= 0.7 \times 10^{- 4} C = 70 μ C$

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A capacitor of capacitance 100 μ F is connected across a battery of emf 6.0 V through a resistance of 20 k Ω for 4.0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0 s after the battery is disconnected ?

c=100μF,Eμf=6V, r=20K,T=4s Charging, Q=CV(1−e−tRC) =6×10−4(1−e−2) =5.187×10−4C Discharging Q=q(e−tRC) =5.184×10−4×e−2 =0.7×10−4C=70μC

A capacitor of capacitance 100 $μ$ F is connected across a battery of emf 6.0 V through a resistance of 20 k $Ω$ for 4.0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0 s after the battery is disconnected ?

$c = 100 μ F, E μ f = 6 V,$

$r = 20 K, T = 4 s$

$Charging,$

$Q = C V (1 - e^{- \frac{t}{R C}})$

$= 6 \times 10^{- 4} (1 - e^{- 2})$

$= 5.187 \times 10^{- 4} C$

$Discharging$

$Q = q (e^{-} \frac{t}{R C})$

$= 5.184 \times 10^{- 4} \times e^{- 2}$

$= 0.7 \times 10^{- 4} C = 70 μ C$