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A $20 \mu \mathrm{F}$ capacitor is connected to $45 \mathrm{~V}$ battery through a circuit whose resistance is $2000 \Omega$. What is the final charge on the capacitor?
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Solution:
2977 Upvotes
Verified Answer
The correct answer is:
$9 \times 10^{-4} \mathrm{C}$
We know that in steady state the capacitor behaves like as an open circuit i.e., capacitor will not pass the current.
So, the potential difference across the capacitor $=45 \mathrm{~V}$
Hence, the final charge on the capacitor is
$$
q=C V
$$
Here, $C=20 \mu \mathrm{F}, V=45 \mathrm{~V}$
$\therefore$
$$
q=20 \times 10^{-6} \times 45
$$
or
$$
q=900 \times 10^{-6}
$$
Or
$$
q=9 \times 10^{-4} \mathrm{C}
$$
So, the potential difference across the capacitor $=45 \mathrm{~V}$
Hence, the final charge on the capacitor is
$$
q=C V
$$
Here, $C=20 \mu \mathrm{F}, V=45 \mathrm{~V}$
$\therefore$
$$
q=20 \times 10^{-6} \times 45
$$
or
$$
q=900 \times 10^{-6}
$$
Or
$$
q=9 \times 10^{-4} \mathrm{C}
$$
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