A 20 - F capacitor is joined to a battery of emf 6-0 V through a resistance of 100 - Find the charge on the capacitor 2-0 ms after the connections are made-

Q = Q.e ^ t/RC = 2 × 10^_3/10 ^2. 20 × 10^ 6 = 120 × 10 ^ 6 e = 120 × 10^ 6 e^ 1 =76 μ C ...

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Byju's Answer

Standard XII

Physics

Charging of Capacitors

A 20 μ F capa...

Question

A 20 $μ F$ capacitor is joined to a battery of emf 6.0 V through a resistance of 100 $Ω$ . Find the charge on the capacitor 2.0 ms after the connections are made.

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Solution

Q = Q. $e^{\frac{- t}{R C}}$

$= \frac{2 \times 10^{_{3}}}{10^{2} . 20 \times 10^{- 6}}$

= $120 \times 10^{- 6} e$

= $120 \times 10^{- 6} e^{- 1}$

= $76 μ C$

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A 20 μF capacitor is joined to a battery of emf 6.0 V through a resistance of 100 Ω. Find the charge on the capacitor 2.0 ms after the connections are made.

Q = Q.e−tRC =2×103102.20×10−6 = 120×10−6e = 120×10−6e−1 =76μC

A 20 $μ F$ capacitor is joined to a battery of emf 6.0 V through a resistance of 100 $Ω$ . Find the charge on the capacitor 2.0 ms after the connections are made.

Q = Q. $e^{\frac{- t}{R C}}$

$= \frac{2 \times 10^{_{3}}}{10^{2} . 20 \times 10^{- 6}}$

= $120 \times 10^{- 6} e$

= $120 \times 10^{- 6} e^{- 1}$

= $76 μ C$