Problem 1 - Applied Physics

Question

For an ac circuit with \(R=1.00 \mathrm{k} \Omega, C=1.00 \mu \mathrm{F}, E=10 \overline{0} \mathrm{~V}\), and \(f=10 \overline{0} \mathrm{~Hz}:\) (a) find the impedance (in ohms), (b) find the phase angle, and (c) find the current.

Step-by-Step Solution

Answer

(a) 1012.4 Ω, (b) -9.06°, (c) 9.88 mA.

Step 1: Find the Reactance of the Capacitor

The reactance of a capacitor can be found using the formula: \( X_C = \frac{1}{2\pi f C} \). Substitute the given values: \( X_C = \frac{1}{2\pi \times 1000 \times 1 \times 10^{-6}} \). Calculate the result: \( X_C = 159.15 \ \Omega \).

Step 2: Calculate the Impedance

The impedance of an R-C circuit is given by: \( Z = \sqrt{R^2 + X_C^2} \). Substitute the given values: \( Z = \sqrt{(1000)^2 + (159.15)^2} \). Calculate the result: \( Z = 1012.4 \ \Omega \).

Step 3: Determine the Phase Angle

The phase angle \( \phi \) can be calculated using: \( \phi = \tan^{-1} \left( \frac{-X_C}{R} \right) \). Substitute the given values: \( \phi = \tan^{-1} \left( \frac{-159.15}{1000} \right) \). Calculate the phase angle: \( \phi = -9.06^\circ \).

Step 4: Calculate the Circuit Current

The current \( I \) in the circuit can be found using Ohm's law: \( I = \frac{E}{Z} \). Substitute the given values: \( I = \frac{10}{1012.4} \). Calculate the current: \( I = 9.88 \times 10^{-3} \text{ A} \).

Key Concepts

Impedance

Impedance in an AC circuit is like the total opposition to the flow of alternating current. It's made up of both resistance, which doesn't change with frequency, and reactance, which varies with frequency. When you look at a circuit with components like resistors and capacitors, you need to calculate the impedance to understand how the circuit will behave. To find the impedance, you use the formula:

• \( Z = \sqrt{R^2 + X_C^2} \)

This equation shows us that impedance combines both the resistance \( R \) and the capacitive reactance \( X_C \). The unit for impedance is Ohms (\( \Omega \)). Remember, impedance takes both magnitude and phase into account, which gives a more complete picture of how alternating current is resisted in a circuit.

Reactance

Reactance is part of the total impedance in AC circuits, representing the opposition due to capacitance or inductance. It is frequency-dependent, meaning it changes with the frequency of the alternating current.For capacitive reactance, the formula used is:

• \( X_C = \frac{1}{2\pi f C} \)

Where:

• \( f \) is the frequency of the AC signal in Hertz (Hz).
• \( C \) is the capacitance in Farads (F).

Capacitive reactance \( X_C \) is measured in Ohms (\( \Omega \)) and decreases as the frequency increases. This makes capacitors great for filtering applications, where you want to block low-frequency signals.By understanding reactance, you can better anticipate how a circuit will react to changes in frequency.

Phase Angle

The phase angle in an AC circuit illustrates the time difference between the voltage and current waveforms. It tells you whether the current leads or lags the voltage.To find the phase angle \( \phi \), we use:

• \( \phi = \tan^{-1} \left( \frac{-X_C}{R} \right) \)

Here:

• \( R \) is the resistance in Ohms.
• \( X_C \) is the capacitive reactance.

If the phase angle is negative, like in our example, it shows the current lags the voltage, typical in RC circuits. Negative angles mean capacitors are in use. The phase angle helps predict how efficiently your circuit can transfer energy.

Ohm's Law

Ohm's Law is a foundational concept in both DC and AC circuits. It relates voltage \( E \), current \( I \), and impedance \( Z \) in an AC circuit. The relationship is given by:

• \( I = \frac{E}{Z} \)

Where:

• \( I \) is the circuit current in Amperes (A).
• \( E \) represents the source voltage in Volts (V).
• \( Z \) is the impedance we previously discussed.

Using Ohm’s Law, you can determine how much current will flow through your AC circuit, based on the voltage applied and the circuit's impedance. This fundamental relationship allows us to analyze and design a variety of electrical circuits with confidence.