MHT CET · Physics · Capacitance
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?
- A \(9 \times 10^{-4} \mathrm{C}\)
- B \(9.154 \times 10^{-4} \mathrm{C}\)
- C \(9.8 \times 10^{-4} \mathrm{C}\)
- D None of these
Answer & Solution
Correct Answer
(A) \(9 \times 10^{-4} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
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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