Problem 64 - University Physics with Modern Physics

Question

An \(L \cdot R-C\) series circuit has \(R=500 \Omega, L=2.00 \mathrm{H}\) \(C=0.500 \mu \mathrm{F},\) and \(V=100 \mathrm{V} .\) (a) For \(\omega=800 \mathrm{rad} / \mathrm{s},\) calculate \(V_{R}, V_{L}, V_{C},\) and \(\phi .\) Using a single set of axes, graph \(v, v_{R}, v_{L},\) and \(v_{C}\) as functions of time. Include two cycles of \(v\) on your graph. (b) Repeat part (a) for \(\omega=1000 \mathrm{rad} / \mathrm{s} .\) (c) Repeat part (a) for \(\omega=1250 \mathrm{rad} / \mathrm{s}\)

Step-by-Step Solution

Answer

Calculate impedance components, voltage drops, phase angle, and plot voltages for each frequency.

Step 1: Understand Series RLC Circuit

An RLC series circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series with a voltage source (V). Each component has a voltage drop across it: \(V_R\) for the resistor, \(V_L\) for the inductor, and \(V_C\) for the capacitor. The goal is to find these voltage drops and the phase angle \(\phi\) for different frequencies \(\omega\).

Step 2: Calculate Impedance Components

The total impedance in the circuit is determined by the resistive, inductive, and capacitive reactances. These are calculated as follows: - Inductive reactance: \(X_L = \omega L\)- Capacitive reactance: \(X_C = 1/(\omega C)\)- Total impedance: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

Step 3: Calculate Voltage Drops

The voltage drop across each component is calculated using Ohm's Law:- \(V_R = I R \),- \(V_L = I X_L \),- \(V_C = I X_C \),where \(I = V/Z\) is the current that flows through the circuit. Insert the values for \(R\), \(L\), \(C\), and \(\omega\) to compute these for each frequency.

Step 4: Calculate Phase Angle (\(\phi\))

The phase angle \(\phi\) shows the delay between the source voltage and the current. It is calculated using \(\tan \phi = \frac{X_L - X_C}{R}\). The angle itself is \(\phi = \arctan\left(\frac{X_L - X_C}{R}\right)\).

Step 5: Graph the Voltages

For graphing on a set of axes, treat each voltage as a sin function, such that:- \(v(t) = V \sin(\omega t)\),- \(v_R(t) = V_R \sin(\omega t)\),- \(v_L(t) = V_L \sin(\omega t + \frac{\pi}{2})\),- \(v_C(t) = V_C \sin(\omega t - \frac{\pi}{2})\).Include two cycles of the source voltage in the graph.

Step 6: Repeat for Different Frequencies

Repeat the calculations in Steps 2 through 5 for \(\omega = 1000 \, \mathrm{rad/s}\) and \(\omega = 1250 \, \mathrm{rad/s}\). Adjust values, calculate new impedance components, voltage drops, phase angles, and re-plot the graphs using each respective frequency.

Key Concepts

Impedance Calculation

In an RLC circuit, impedance is a crucial concept as it determines how the current behaves in the circuit. Impedance combines resistance (R), inductive reactance \(X_L\), and capacitive reactance \(X_C\) into a single measure of opposition against the flow of alternating current.

To calculate the impedance in an RLC series circuit, we use the formula:

• Inductive Reactance: \(X_L = \omega L\)
• Capacitive Reactance: \(X_C = \frac{1}{\omega C}\)
• Total Impedance: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\)

You can think of impedance \(Z\) as the total 'ohmic resistance' to the flow of current, but more complex, since it takes into account the frequency \(\omega\) of the source voltage. This frequency affects both the inductive and capacitive parts of the circuit because reactance changes with frequency. Higher frequencies will increase \(X_L\) and decrease \(X_C\), impacting the total impedance. By using the formula for \(Z\), you can calculate the opposition for different frequencies.

Voltage Drops in RLC Circuits

In RLC circuits, understanding voltage drops across each component helps in analyzing the circuit behavior. Each component (resistor, inductor, capacitor) has a voltage drop because they impede current flow in different ways.

For calculating voltage drops:

• Use Ohm's law: \(V = IZ\) where \(I\) is the current.
• Current \(I\) is given by \(I = \frac{V}{Z}\), with \(V\) as the supply voltage and \(Z\) as the calculated impedance.
• Resistor Voltage Drop: \(V_R = IR\)
• Inductor Voltage Drop: \(V_L = IX_L\)
• Capacitor Voltage Drop: \(V_C = IX_C\)

Voltage across the resistor \(V_R\) is in-phase with the current, whereas \(V_L\) leads the current and \(V_C\) lags the current by 90 degrees. This phase shift is key to understanding how these components react differently in an AC circuit. It's essential to calculate these drops at different frequencies as each will alter the reactance values, changing \(I\) and thus affecting \(V_R, V_L,\) and \(V_C\).

Phase Angle in RLC Circuits

The phase angle \(\phi\) in an RLC circuit indicates the difference in timing between the voltage across the entire circuit and the current running through it. This phase shift arises due to the reactive components (inductor and capacitor).

To determine the phase angle, the formula used is:

• \(\tan \phi = \frac{X_L - X_C}{R}\)
• Thus, \(\phi = \arctan\left(\frac{X_L - X_C}{R}\right)\)

The phase angle is important because it informs how much the current leads or lags the voltage. When \(X_L > X_C\), the voltage leads the current, creating an inductive circuit. Conversely, when \(X_C > X_L\), the current leads, producing a capacitive circuit. At resonance (where \(X_L = X_C\)), the phase angle is zero, meaning the voltage and current are in phase. Adjusting frequencies affects \(X_L\) and \(X_C\), thus changing the phase angle and the overall response of the circuit. Understanding this can help you predict and manage circuit behavior under different operating conditions.