Question

An alternating voltage $E = 200\sqrt{2} \sin(100t)$ volt is connected to a $1 \mu\text{F}$ capacitor through an a.c. ammeter. The reading of the ammeter shall}

• A. 10 mA
• B. 20 mA
• C. 40 mA
• D. 80 mA

Hint:

A.C. ammeters always measure the R.M.S. value of the current, which is $I_{rms} = V_{rms} / X_C$.

Answer 1

The Correct Option is A

Step 1: From the given voltage equation $E = 200\sqrt{2} \sin(100t)$, compare with the standard form $E = E_0 \sin(\omega t)$ to find the peak voltage $E_0 = 200\sqrt{2}$ V and angular frequency $\omega = 100$ rad/s.
Step 2: The root mean square (r.m.s.) voltage $V_{rms}$ measured by an a.c. meter is \[ V_{rms} = \frac{E_0}{\sqrt{2 = \frac{200\sqrt{2{\sqrt{2 = 200 \text{ V} \]
Step 3: Calculate the capacitive reactance $X_C$ for a capacitance $C = 1 \mu\text{F} = 1 \times 10^{-6}$ F: \[ X_C = \frac{1}{\omega C} = \frac{1}{100 \times 1 \times 10^{-6 = 10^4 \Omega \]
Step 4: The ammeter reading is the r.m.s. current $I_{rms}$: \[ I_{rms} = \frac{V_{rms{X_C} = \frac{200}{10^4} = 0.02 \text{ A} = 20 \text{ mA} \]