Question 298: AC fundamentals

Question

An alternating voltage $\mathrm{E}=100 \sqrt{2} \sin (50 \mathrm{t})$ is connected to a $2 \mu \mathrm{~F}$ capacitor through an a.c. ammeter. The ammeter reading will be

Accepted Answer

Given:
Alternating voltage: $ E = 100\sqrt{2} \sin(50t) $
Capacitance: $ C = 2\,\mu	ext{F} = 2 	imes 10^{-6}\,	ext{F} $
Let us find the ammeter reading (the
rms current
through the capacitor).
Step 1: Compare with Standard AC Voltage Equation
The general equation for AC voltage is
$ E = E_0 \sin \omega t $
Here, $ E_0 = 100\sqrt{2} $, $ \omega = 50 $ rad/s.
Step 2: Capacitive Reactance ($ X_C $)
$ X_C = \frac{1}{\omega C} $
Substitute the values:
$ X_C = \frac{1}{50 	imes 2 	imes 10^{-6}}     = \frac{1}{1 	imes 10^{-4}}     = 10,000~\Omega $
Step 3: Find Peak Current ($ I_0 $)
$ I_0 = \frac{E_0}{X_C} $
Substitute the values:
$ I_0 = \frac{100\sqrt{2}}{10,000} = \frac{100 	imes 1.414}{10,000} = \frac{141.4}{10,000} = 0.01414\,	ext{A} = 14.14\,	ext{mA} $
Step 4: Find RMS Current ($ I_{	ext{rms}} $)
$ I_{	ext{rms}} = \frac{I_0}{\sqrt{2}} $
Substitute the value of $ I_0 $:
$ I_{	ext{rms}} = \frac{14.14~	ext{mA}}{1.414} = 10~	ext{mA} $
Final Answer:
$ \boxed{10~	ext{mA}} $
(Option A)

Answer

10 mA

Answer

5 mA

Answer

20 mA

Answer

30 mA

AC fundamentals

An alternating voltage \mathrm{E}=100 \sqrt{2} \sin (50 \mathrm{t}) is connected to a 2 \mu \mathrm{~F} capacitor through an a.c. ammeter. The ammeter reading will be

Choose the correct answer: