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A \(105-\mu F\) capacitor is connected to an ac generator with an rms voltage of \(20.0 \mathrm{V}\) and a frequency of \(100.0 \mathrm{Hz}\). What is the rms current in this circuit?

Short Answer

Expert verified
The rms current is approximately 1.32 A.

Step by step solution

01

Understand the Formula for Capacitive Reactance

First, we need to determine the capacitive reactance with the formula: \[ X_C = \frac{1}{2\pi f C} \], where:- \( f \) is the frequency of the generator (\(100.0\, \text{Hz}\)), and- \( C \) is the capacitance (\(105\, \mu\text{F} = 105 \times 10^{-6} \text{F}\)).
02

Calculate the Capacitive Reactance

Calculate the capacitive reactance \( X_C \):\[ X_C = \frac{1}{2\pi \times 100.0 \times 105 \times 10^{-6}}\]\[ X_C \approx \frac{1}{0.066025} \approx 15.14 \Omega \].
03

Use Ohm’s Law for AC Circuits

Now that we have the capacitive reactance, we can find the rms current using Ohm’s Law for AC circuits: \[ I_{rms} = \frac{V_{rms}}{X_C} \], where \( V_{rms} = 20.0\,\text{V} \) and \( X_C \approx 15.14\, \Omega \).
04

Calculate the RMS Current

Substitute the known values into the formula to find \( I_{rms} \):\[ I_{rms} = \frac{20.0}{15.14} \approx 1.32 \text{A} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitive Reactance
In an AC circuit, capacitive reactance is a crucial concept when dealing with capacitors. It represents how much a capacitor resists the flow of alternating current. Unlike pure resistance found in resistors, capacitive reactance is dependent on two key variables: the frequency of the AC signal and the capacitance of the capacitor itself. The formula used to calculate capacitive reactance is:
  • \( X_C = \frac{1}{2\pi f C} \)
Here, \( X_C \) denotes the capacitive reactance, \( f \) stands for the frequency in hertz, and \( C \) is the capacitance in farads.

Note that as the frequency of the AC signal increases, the capacitive reactance decreases, making it easier for the current to flow. Conversely, if the capacitance increases, the reactance decreases as well. By understanding this concept, you can better predict how a capacitor will behave in an AC circuit.
Ohm's Law for AC Circuits
Ohm's Law is a fundamental principle in electrical engineering, and it applies to both DC and AC circuits. However, in AC circuits that involve capacitive or inductive elements, we use a modified version of Ohm’s Law due to the presence of reactance. The formula is given by:
  • \( I_{rms} = \frac{V_{rms}}{X_C} \)
Here, \( I_{rms} \) is the root mean square current, \( V_{rms} \) is the root mean square voltage, and \( X_C \) is the capacitive reactance.

Ohm's Law for AC circuits helps us to understand how the voltage and current interact within circuits containing capacitive elements. By using this formula, you can determine how much current flows through a capacitor when a certain AC voltage is applied. It also illustrates the inverse relationship between capacitive reactance and current: as the capacitive reactance decreases, the current increases.
RMS Current
When working with AC circuits, we often refer to the root mean square (RMS) current. RMS is a statistical measure used to represent the effective value of an alternating current. It provides a way to compare AC currents to DC currents in terms of the power they deliver.
  • The formula for RMS current is derived using the following relationship: \( I_{rms} = \frac{V_{rms}}{X_C} \), where \( V_{rms} \) is the voltage and \( X_C \) is the capacitive reactance.
Estimating RMS current is essential for understanding how a circuit operates in real-world conditions, as it reflects the true power that can be utilized by the circuit's components.

In simpler terms, RMS current is an indication of the equivalent DC current value that would yield the same power dissipation in a resistive load. This makes it much easier to visualize and calculate power usage in AC systems. Understanding RMS values is fundamental when analyzing AC circuit behavior, helping you to optimize circuit designs effectively.

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Most popular questions from this chapter

CE Predict/Explain When a long copper wire of finite resistance is connected to an ac generator, as shown in Figure \(24-28\) (a). a certain amount of current flows through the wire. The wire is now wound into a coil of many loops and reconnected to the generator, as indicated in Figure \(24-28\) (b). (a) Is the current supplied to the coil greater than, less than, or the same as the current supplied to the uncoiled wire? (b) Choose the best explanation from among the following: I. More current flows in the circuit because the coiled wire is an inductor, and inductors tend to keep the current flowing in an ac circuit. II. The current supplied to the circuit is the same because the wire is the same. Simply wrapping the wire in a coil changes nothing. III. Less current is supplied to the circuit because the coiled wire acts as an inductor, which increases the impedance of the circuit.

CE The voltage in a sinusoidally driven \(R L C\) circuit leads the current. (a) If we want to bring this circuit into resonance by changing the frequency of the generator, should the frequency be increased or decreased? Explain. (b) If we want to bring this circuit into resonance by changing the inductance instead, should the inductance be increased or decreased? Explain.

CE BIO Persistence of Vision Although an incandescent lightbulb appears to shine with constant intensity, this is an artifact of the eye's persistence of vision. In fact, the intensity of a bulb's light rises and falls with time due to the alternating current used in household circuits. If you could perceive these oscillations, would you see the light attain maximum brightness 60 or 120 times per second? Explain.

IP An RLC circuit has a resistance of \(105 \Omega\), an inductance of \(85.0 \mathrm{mH},\) and a capacitance of \(13.2 \mu \mathrm{F} .\) (a) What is the power factor for this circuit when it is connected to a \(125-\mathrm{Hz}\) ac generator? (b) Will the power factor increase, decrease, or stay the same if the resistance is increased? Explain. (c) Calculate the power factor for a resistance of \(525 \Omega\)

P When a certain resistor is connected to an ac generator with a maximum voltage of \(15 \mathrm{V},\) the average power dissipated in the resistor is 22 W. (a) What is the resistance of the resistor? (b) What is the rms current in the circuit? (c) We know that \(P_{\mathrm{ar}}=I_{\mathrm{rms}}^{2} R,\) and hence it seems that reducing the resistance should reduce the average power. On the other hand, we also know that \(P_{a v}=V_{\text {rms }}^{2} / R,\) which suggests that reducing \(R\) increases \(P_{a v}\). Which conclusion is correct? Explain.

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