Chapter 10: Problem 54
An alternating voltage \(E=200 \sqrt{2} \sin (100 t) \mathrm{V}\) is connected to a \(1 \mathrm{mF}\) capacitor through an ac ammeter. What will be the reading of the ammeter? a. \(40 \mathrm{~mA}\) b \(20 \mathrm{~mA}\) c. \(30 \mathrm{~m} \mathrm{~A}\) d. \(10 / \sqrt{2} \mathrm{~m} \mathrm{~A}\)
Short Answer
Step by step solution
Understand the Voltage Function
Identify Capacitive Reactance
Calculate Root Mean Square (RMS) Voltage
Use Ohm’s Law for AC Circuits
Identify the Ammeter Reading
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Alternating Current
Capacitive Reactance
- It is inversely proportional to the frequency of the AC signal and the capacitance of the capacitor (\(X_C = \frac{1}{\omega C}\)).
- Higher frequencies result in lower reactance, allowing more current to pass through the capacitor.
- In our exercise, using a capacitance \(C = 1 \mathrm{mF}\) and frequency \(\omega = 100\ \text{rad/s}\), \(X_C\) is calculated to be \(10 \ \Omega\).
Ohm's Law for AC Circuits
- \(I_{rms}\) is the root mean square (RMS) current.
- \(V_{rms}\) is the RMS voltage.
- \(X_C\) is the capacitive reactance.
RMS Voltage
- It is calculated as \(V_{rms} = \frac{V_{peak}}{\sqrt{2}}\) for sinusoidal waves.
- Gives the equivalent DC value which provides the same power to a load.
- In the exercise, converting the supplied peak voltage \(200\sqrt{2}\) volts to RMS results in \(200\ \text{V}\).