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GUJCET · Physics · Alternating Current
An alternating voltage given as \(\mathbf{V}=100 \sqrt{2} \sin 100 t\) Volt is applied to a capacitor of \(1 \mu \mathrm{~F}\). The current reading of the ammeter will be equal to ____________ mA .
- A 10
- B 20
- C 40
- D 80
Answer & Solution
Correct Answer
(A) 10
Step-by-step Solution
Detailed explanation
(A) 10
\(I=\frac{V}{Z}\)
\(\mathrm{I}=\frac{v_{m}}{\sqrt{2} \times \frac{1}{\omega \mathrm{C}}}\)
\(\mathrm{I}=\frac{v_{m} \times \omega \mathrm{C}}{\sqrt{2}}\)
\(I=\frac{(100 \sqrt{2}) \times(100) \times\left(1 \times 10^{-6}\right)}{\sqrt{2}}\)
\(\mathrm{I}=10^{-2}\)
\(\mathrm{I}=10 \times 10^{-3} \mathrm{~A}\)
\(\mathrm{I}=10 \mathrm{~mA}\)
\(I=\frac{V}{Z}\)
\(\mathrm{I}=\frac{v_{m}}{\sqrt{2} \times \frac{1}{\omega \mathrm{C}}}\)
\(\mathrm{I}=\frac{v_{m} \times \omega \mathrm{C}}{\sqrt{2}}\)
\(I=\frac{(100 \sqrt{2}) \times(100) \times\left(1 \times 10^{-6}\right)}{\sqrt{2}}\)
\(\mathrm{I}=10^{-2}\)
\(\mathrm{I}=10 \times 10^{-3} \mathrm{~A}\)
\(\mathrm{I}=10 \mathrm{~mA}\)
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