Problem 16 - University Physics with Modern Physics

Question

\(\mathrm{A} 200-\Omega\) resistor, a \(0.900-\mathrm{H}\) inductor, and a \(6.00-\mu \mathrm{F}\) capacitor are connected in series across a voltage source that has voltage amplitude 30.0 \(\mathrm{V}\) and an angular frequency of 250 \(\mathrm{rad} / \mathrm{s}\) . (a) What are \(v, v_{R}, v_{L},\) and \(v_{C}\) at \(t=20.0 \mathrm{ms}\) ? Compare \(v_{R}+\) \(v_{L}+v_{C}\) to \(v\) at this instant. (b) What are \(V_{R}, V_{L},\) and \(V_{C} ?\) Compare \(V\) to \(V_{R}+V_{L}+V_{C} .\) Explain why these two quantities are not equal.

Step-by-Step Solution

Answer

The combined voltages are out of phase, thus the total peak voltage seems greater than the source. For peaks: \( V_R + V_L + V_C > V \). At \( t=20ms \), they match the source \( v \).

Step 1: Calculate the impedance and individual voltages

First, calculate the impedances for the resistor, inductor, and capacitor. The resistance is \( R = 200 \, \Omega \). The inductive reactance is given by \( X_L = \omega L = 250 \, \text{rad/s} \times 0.900 \, \text{H} = 225 \; \Omega \). The capacitive reactance is \( X_C = \frac{1}{\omega C} = \frac{1}{250 \, \text{rad/s} \times 6.00 \times 10^{-6} \, \text{F}} = 666.67 \, \Omega \).

Step 2: Calculate the total impedance

The total impedance \( Z \) can be calculated using the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Substituting the values we have: \( Z = \sqrt{200^2 + (225 - 666.67)^2} \approx 546.53 \, \Omega \).

Step 3: Determine the peak current and voltages

The peak current \( I_0 \) can be calculated by dividing the peak voltage \( V_0 = 30 \, \text{V} \) by the total impedance \( Z \). So, \( I_0 = \frac{30}{546.53} \approx 0.0549 \, \text{A} \). The peak voltages across the resistor, inductor, and capacitor are: \( V_R = I_0 \times R = 0.0549 \times 200 \approx 11.0 \, \text{V} \), \( V_L = I_0 \times X_L = 0.0549 \times 225 \approx 12.35 \, \text{V} \), and \( V_C = I_0 \times X_C = 0.0549 \times 666.67 \approx 36.6 \, \text{V} \).

Step 4: Evaluate voltage at time t=20ms

The instantaneous voltage across the circuit at time \( t = 20 \, \text{ms} \) is \( v(t) = V_0 \cos(\omega t) = 30 \cos(250 \times 0.020) \). Similarly, calculate \( v_R, v_L, \) and \( v_C \) using the formula \( i(t) = I_0 \cos(\omega t) \), then \( v_R = i(t) \times R \), \( v_L = i(t) \times X_L \sin(\omega t) \), and \( v_C = i(t) \times X_C \sin(\omega t) \).

Step 5: Compare sum of instantaneous voltages to source voltage

Using the values from Step 4, calculate \( v_R + v_L + v_C \) and compare it with \( v \) at \( t = 20 \, \text{ms} \). Due to phase differences in the voltages, \( v_R + v_L + v_C \) should equal \( v \). In ideal calculations, they should match.

Step 6: Summarize peak voltage comparison

The apparent sum of the individual voltages \( V_R + V_L + V_C \) is greater than the source voltage \( V \). This occurs because the voltages are out of phase, meaning they do not reach their peaks simultaneously.

Step 7: Explain phase relationships

In an AC circuit, the out-of-phase nature of voltages across different circuit components leads to additive effects being countered when they are summed over waves with phases. The phasor addition principle explains this discrepancy between the simple arithmetic sum of peak voltages and the actual source RMS voltage.

Key Concepts

Impedance Calculation

To effectively analyze AC circuits, understanding impedance is crucial. Impedance, symbolized as \( Z \), is the total opposition that a circuit presents to the flow of alternating current. It combines the effects of resistance, inductive reactance, and capacitive reactance.

Resistance \( R \) is straightforward, calculated simply as \( 200 \, \Omega \) in this case. For the inductor, the inductive reactance \( X_L \) is calculated using the formula \( X_L = \omega L \), where \( \omega \) is the angular frequency (250 rad/s), and \( L \) is the inductance (0.900 H). This results in \( 225 \, \Omega \) of reactance.

The capacitive reactance \( X_C \) is determined using \( X_C = \frac{1}{\omega C} \), resulting in \( 666.67 \, \Omega \). Finally, the total impedance is found using the relationship:

• \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)

This calculation gives us an impedance of approximately \( 546.53 \, \Omega \). Impedance is crucial as it not only affects how much current flows through the circuit but also influences the phase relationship between voltage and current.

AC Circuit

Alternating current (AC) circuits operate with currents and voltages that vary with time, most commonly in a sinusoidal manner. In our example, the voltage source has an amplitude of 30 V with an angular frequency of 250 rad/s. AC circuits include resistors, inductors, and capacitors, each contributing differently to the circuit's overall behavior.

In AC circuits:

• Resistors maintain the same basis of operation, obeying Ohm’s Law where voltage and current are in phase.
• Inductors cause the current to lag behind the voltage due to the opposing voltage generated when the current changes.
• Capacitors cause the current to lead the voltage, as they can source current even while the voltage is still building up.

These elements interact to change the phase relationships, which can significantly affect the performance of the circuit. In this specific setup, the components are in series and the source voltage provides the periodic driving force through this configuration.

Phasor Addition

When analyzing AC signal relationships, phasor addition helps understand how voltages across components lead to an overall circuit function. Unlike DC circuits, AC voltages across series elements do not simply add up arithmetically due to their phase differences.

Phasors are a representation of sinusoidal functions as rotating vectors in the complex plane, which allow for simple addition and subtraction of waveforms by considering their magnitude and phase angle. For example, in our exercise, the peak voltages across the resistor \( V_R \), inductor \( V_L \), and capacitor \( V_C \) were calculated by multiplying their respective reactances by the peak current. However, their sum is not directly the source voltage’s peak.

Due to phase-shifted contributions:

• The phasor sum \( V_R + V_L + V_C \) exceeds the simple algebraic sum when treated without phase considerations.
• The difference represents phase shifts — the resistor voltage is in phase with the current, while the inductor and capacitor voltages are shifted, which leads to complex interaction between them.

This reveals why the arithmetic sum of these peak values is larger than the actual root-mean-square (RMS) voltage of the circuit, illustrating the importance of using phasor addition for accurate AC circuit analysis.