MHT CET · Physics · Alternating Current
An alternating voltage \(v=200 \sqrt{2} \sin (100 t)\) is connected to a \(1 \mu \mathrm{~F}\) capacitor through an a. c. ammeter. The reading of the ammeter shall be
- A 10 mA
- B 20 mA
- C 40 mA
- D 80 mA
Answer & Solution
Correct Answer
(B) 20 mA
Step-by-step Solution
Detailed explanation
Alternating voltage: \(\mathrm{e}=200 \sqrt{2} \sin (100 \mathrm{t})\) volt Comparing with \(\mathrm{e}=\mathrm{e}_0 \sin \omega t\)
\(\omega=100 \mathrm{rad} / \mathrm{s}, \mathrm{e}_0=200 \sqrt{2}\)
Capacitive reactance,
\(\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10^{-6}} \Omega=10^4 \Omega \)
\( \mathrm{I}_0=\frac{\mathrm{e}_0}{\mathrm{X}_{\mathrm{C}}} \)
\( \mathrm{I}_0=\frac{200 \sqrt{2}}{10^4} \)
\( \mathrm{I}_0=2 \sqrt{2} \times 10^{-2} \mathrm{~A} \)
\( \mathrm{I}_{\mathrm{rmi}}=\frac{\mathrm{I}_0}{\sqrt{2}}=\frac{2 \sqrt{2} \times 10^{-2}}{\sqrt{2}}=2 \times 10^{-2}~ \cdot\) \(\mathrm{~A}=20 \mathrm{~mA}\)
\(\omega=100 \mathrm{rad} / \mathrm{s}, \mathrm{e}_0=200 \sqrt{2}\)
Capacitive reactance,
\(\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10^{-6}} \Omega=10^4 \Omega \)
\( \mathrm{I}_0=\frac{\mathrm{e}_0}{\mathrm{X}_{\mathrm{C}}} \)
\( \mathrm{I}_0=\frac{200 \sqrt{2}}{10^4} \)
\( \mathrm{I}_0=2 \sqrt{2} \times 10^{-2} \mathrm{~A} \)
\( \mathrm{I}_{\mathrm{rmi}}=\frac{\mathrm{I}_0}{\sqrt{2}}=\frac{2 \sqrt{2} \times 10^{-2}}{\sqrt{2}}=2 \times 10^{-2}~ \cdot\) \(\mathrm{~A}=20 \mathrm{~mA}\)
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