Chapter 22: Problem 7
The voltage amplitude of an ac source is \(25.0 \mathrm{~V},\) and its angular frequency is \(1000 \mathrm{rad} / \mathrm{s}\). Find the current amplitude if the capacitance of a capacitor connected across the source is (a) \(0.0100 \mu \mathrm{F}\) (b) \(1.00 \mu \mathrm{F}\), (c) \(100 \mu \mathrm{F}\).
Short Answer
Step by step solution
Identify the Given Values
Formula for Capacitive Reactance
Calculate the Current Amplitude
Solve for each Capacitance (Part a)
Solve for each Capacitance (Part b)
Solve for each Capacitance (Part c)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Voltage Amplitude
In our example, the voltage amplitude is given as 25.0 V. This means that at the peak, the AC wave can push with a potential difference of 25 Volts. This value is crucial when calculating other properties such as current amplitude, especially when we consider how voltages translate to currents through elements like resistors or capacitors, which the AC wave will interact with at different phases of its cycle.
Understanding voltage amplitude allows us to predict the behavior of the electrical devices connected to the source, like their power consumption or how they perform under varying conditions.
Angular Frequency
Our given angular frequency is 1000 rad/s. This rapid rate of oscillation signifies a high frequency AC source, which can lead to less time between each cycle of voltage and current change. Calculations involving angular frequency, like those leading to the current amplitude, often involve it in radians because it’s part of the circular movement defining wave motion.
We calculate parameters such as capacitive reactance using this frequency, helping us understand the rapid changes in electrical systems. Thus, knowing the angular frequency assists in assessing how fast the entire circuit is reacting to those changes.
Capacitive Reactance
The formula for capacitive reactance is:\[ X_C = \frac{1}{\omega C} \]where
- \(X_C\) stands for capacitive reactance,
- \(\omega\) is the angular frequency,
- and \(C\) is the capacitance.
In the given exercise, different capacitance values yield different reactance, demonstrating how a capacitor's behavior changes with frequency. Lower reactance at high frequencies allows more current to pass, correlating with higher current amplitudes as seen in parts (a) to (c) of the solution.
Current Amplitude
The formula to calculate current amplitude is:\[ I_0 = \frac{V_0}{X_C} \]where
- \(I_0\) is the current amplitude,
- \(V_0\) is the voltage amplitude,
- and \(X_C\) is the capacitive reactance.
In the example, we see how varying the capacitance impacts current amplitude due to fluctuating reactance, demonstrating the delicate balancing act within AC circuits. Understanding this concept helps explain why devices operate differently under alternate frequencies and with various capacitor types.