Problem 5 - Applied Physics

Question

For an ac circuit with resistance \(10.0 \mathrm{~m} \Omega\), capacitance \(5.00 \mu \mathrm{F}\), frequency \(10.0 \mathrm{kHz}\), and \(E=15.0 \mathrm{mV}\) : (a) find the impedance (in ohms), (b) find the phase angle, and (c) find the current.

Step-by-Step Solution

Answer

Impedance is 3.18 Ω, phase angle is 89.8°, and current is 4.72 mA.

Step 1: Calculate Reactance of the Capacitor

The capacitive reactance \( X_c \) is given by the formula \( X_c = \frac{1}{2\pi f C} \), where \( f \) is the frequency and \( C \) is the capacitance. Substitute \( f = 10,000 \) Hz and \( C = 5.0 \times 10^{-6} \) F into the formula:\[ X_c = \frac{1}{2\pi \times 10,000 \times 5.0 \times 10^{-6}} \approx 3.18 \ \Omega \]

Step 2: Calculate Impedance

Impedance \( Z \) in an AC circuit with resistance \( R \) and reactance \( X_c \) is calculated as \( Z = \sqrt{R^2 + X_c^2} \). Substitute \( R = 10.0 \times 10^{-3} \) Ohms and \( X_c = 3.18 \) Ohms into the formula:\[ Z = \sqrt{(10.0 \times 10^{-3})^2 + (3.18)^2} \approx 3.18 \ \Omega \]

Step 3: Calculate Phase Angle

The phase angle \( \phi \) is calculated using the tangent function: \( \tan \phi = \frac{X_c}{R} \). Substitute \( X_c = 3.18 \) Ohms and \( R = 10.0 \times 10^{-3} \) Ohms:\[ \tan \phi = \frac{3.18}{10.0 \times 10^{-3}} \Rightarrow \phi = \tan^{-1}(318) \approx 89.8^\circ \]

Step 4: Calculate Current

The current \( I \) in the circuit is given by Ohm’s law for AC circuits: \( I = \frac{E}{Z} \), where \( E = 15.0 \) mV and \( Z = 3.18 \) Ohms. Substituting these values gives:\[ I = \frac{0.015}{3.18} \approx 4.72 \times 10^{-3} \ \text{A} \] or \( 4.72 \ \text{mA} \).

Key Concepts

Impedance Calculation

In AC circuits, impedance is a measure of opposition that a circuit presents to the passage of alternating current. It combines resistance with the reactive elements, represented as the total opposition within a circuit. To find the impedance (denoted as \( Z \)) in a circuit containing a resistor and a capacitor, you need to understand both resistance (\( R \)) and capacitive reactance (\( X_c \)). Impedance is calculated using the formula: \[ Z = \sqrt{R^2 + X_c^2} \] Here's a quick explanation of the terms:

• Resistance (\( R \)) is the inherent opposition in the circuit due to resistors. It's measured in ohms (\( \Omega \)).
• Capacitive reactance (\( X_c \)) is the opposition to the change of voltage across an element. It affects the phase difference between the voltage and current. Calculated as \( X_c = \frac{1}{2\pi fC} \), where \( f \) is the frequency and \( C \) is the capacitance.

Substituting the given values of \( R = 10.0 \) m\( \Omega \) and \( X_c = 3.18 \) \( \Omega \) into the equation provides an impedance of approximately 3.18 \( \Omega \). This tells us how much overall resistance the AC finds in the circuit.

Phase Angle Determination

The phase angle (denoted as \( \phi \)) in an AC circuit indicates the shift in phase between the voltage across and the current through the network. It's a crucial factor that demonstrates the timing difference and helps in understanding the power factor of the AC circuit. The phase angle is determined by the relative sizes of the capacitive reactance and the resistance in the circuit. You can calculate the phase angle using the following relationship:\[ \tan \phi = \frac{X_c}{R} \]This formula links the phase angle to the reactance and resistance of the circuit:

• \( X_c \) is the reactance offered due to the capacitor.
• \( R \) is the resistance presented by the resistor.

In our example, substituting \( X_c = 3.18 \) \( \Omega \) and \( R = 10.0 \) m\( \Omega \) yields a very large tangent value, leading to a near-90-degree angle. Such a large value occurs because the reactance considerably outmatches the resistance, indicating that the circuit behaves more like a capacitive circuit. This significant phase shift is typical in scenarios where capacitive reactance is several times higher than the resistance.

Current Calculation

The calculation of the current in an AC circuit is similar to that in a DC circuit, but here impedance replaces resistance. AC Ohm's law expresses this relationship simply:\[ I = \frac{E}{Z} \]Where:

• \( I \) is the current through the circuit.
• \( E \) is the source voltage.
• \( Z \) is the impedance we calculated earlier.

Using these values from the exercise, with \( E = 15.0 \) mV and \( Z = 3.18 \) \( \Omega \), substituting gives:\[ I = \frac{0.015}{3.18} \approx 4.72 \times 10^{-3} \]\( \text{A} \),or 4.72 mA. The calculation indicates the flow of current based on the given voltage and the impedance in the circuit. Keep in mind that higher impedance or lower voltage would reduce the current level, and vice versa. This understanding helps in correctly designing AC circuits to achieve the desired performance.