An alternating voltage source \( \upsilon = 100\sqrt{2}\sin(100t)\;V \) is connected to a 1μF capacitor through an ammeter. The reading of the ammeter is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10−2 A
An ac ammeter reads the rms current. For a pure capacitor the current is the voltage divided by the capacitive reactance \( X_C = 1/\omega C \). Comparing with \( \upsilon = V_0\sin\omega t \):
\( V_0 = 100\sqrt{2}\,\mathrm{V} \)
\( \omega = 100\,\mathrm{rad\,s^{-1}} \)
\( C = 1\,\mu\mathrm{F} = 1\times10^{-6}\,\mathrm{F} \)
\( \displaystyle V_{\mathrm{rms}} = \frac{V_0}{\sqrt{2}} = 100\,\mathrm{V} \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{100\,\mathrm{rad\,s^{-1}} \times 1\times10^{-6}\,\mathrm{F}} = 1\times10^{4}\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_C} = \frac{100\,\mathrm{V}}{1\times10^{4}\,\Omega} \)
\( I_{\mathrm{rms}} = 10^{-2}\,\mathrm{A} \)