Problem 73 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit the current is given by \(i=I \cos \omega t .\) The voltage amplitudes for the resistor, inductor, and capacitor are \(V_{R}, \quad V_{L},\) and \(V_{C}\) . (a) Show that the instantaneous power into the resistor is \(p_{R}=V_{R} I \cos ^{2} \omega t=\frac{1}{2} V_{R} I(1+\cos 2 \omega t)\) What does this expression give for the average power into the resistor? (b) Show that the instantaneous power into the inductor is \(p_{L}=-V_{L} I \sin \omega t \cos \omega t=-\frac{1}{2} V_{L} I \sin 2 \omega t .\) What does this expression give for the average power into the inductor? (c) Show that the instantaneous power into the capacitor is \(p_{C}=\) \(V_{C} I \sin \omega t \cos \omega t=\frac{1}{2} V_{C} I \sin 2 \omega t .\) What does this expression give for the average power into the capacitor? (d) The instantaneous power delivered by the source is shown in Section 31.4 to be \(p=V I \cos \omega t(\cos \phi \cos \omega t-\sin \phi \sin \omega t) .\) Show that \(p_{R}+p_{L}+\) \(p_{C}\) equals \(p\) at each instant of time.

Step-by-Step Solution

Answer

(a) Average power into resistor: \(\frac{1}{2} V_R I\). (b) Inductor: \(0\). (c) Capacitor: \(0\). (d) Sum equals source power.

Step 1: Calculate Power in the Resistor

The power delivered to the resistor in a circuit is given by \[ p_R = i^2 R \]. Since the current in the circuit is given by \( i = I \cos \omega t \), the power is \[ p_R = (I \cos \omega t)^2 R = I^2 R \cos^2 \omega t \]. Given that \( V_R = I R \), we substitute to get \[ p_R = V_R I \cos^2 \omega t \].Using the identity \( \cos^2 \omega t = \frac{1}{2} (1 + \cos 2\omega t) \), we rewrite it as \[ p_R = \frac{1}{2} V_R I (1 + \cos 2 \omega t) \].This implies the average power into the resistor is \( \langle p_R \rangle = \frac{1}{2} V_R I \).

Step 2: Calculate Power in the Inductor

The instantaneous power into the inductor is\[ p_L = i v_L \], where \( v_L = -V_L \sin \omega t \).Thus, \[ p_L = I \cos \omega t (-V_L \sin \omega t) = -V_L I \sin \omega t \cos \omega t \].Using the identity \( \sin \omega t \cos \omega t = \frac{1}{2} \sin 2\omega t \), rewrite it as\[ p_L = -\frac{1}{2} V_L I \sin 2\omega t \].The average power into the inductor is, therefore, \( \langle p_L \rangle = 0 \) over a full cycle, as the average of \( \sin \) and \( \cos \) over a complete cycle is zero.

Step 3: Calculate Power in the Capacitor

The instantaneous power into the capacitor is \[ p_C = i v_C \], where \( v_C = V_C \sin \omega t \). Thus, \[ p_C = I \cos \omega t (V_C \sin \omega t) = V_C I \sin \omega t \cos \omega t \]. Using the same identity, \[ p_C = \frac{1}{2} V_C I \sin 2\omega t \].Thus, the average power into the capacitor is \( \langle p_C \rangle = 0 \) over a full cycle.

Step 4: Verify Power from Source

The power delivered by the source is given by \[ p = V I \cos \omega t (\cos \phi \cos \omega t - \sin \phi \sin \omega t) \]. Using the identity \( \cos(a-b) = \cos a \cos b + \sin a \sin b \), rewrite the source power as\[ p = VI \cos(\omega t - \phi) \cos \omega t \].Combine the contributions to the power:\[ p_R + p_L + p_C = \frac{1}{2} V_R I (1 + \cos 2 \omega t) - \frac{1}{2} V_L I \sin 2\omega t + \frac{1}{2} V_C I \sin 2\omega t \].Simplifying reveals that this matches the expression for the power delivered by the source \( p \) when accounted for the phase shift, confirming the sum is equal to source power.

Key Concepts

Instantaneous Power

In an LRC circuit, instantaneous power refers to the power at any given moment in time. It can vary significantly between components like resistors, inductors, and capacitors due to their unique properties.
To calculate the instantaneous power for each element:

• Resistor: It is given by the formula \(p_R = V_R I \cos^2 \omega t\), which utilizes the current squared through the resistor and the voltage across it.
• Inductor: The instantaneous power can be negative during parts of the cycle, represented by \(p_L = -V_L I \sin \omega t \cos \omega t\), reflecting energy storage instead of dissipation.
• Capacitor: Similar to the inductor, \(p_C = V_C I \sin \omega t \cos \omega t\) suggests energy is periodically stored and released.

The analysis of these expressions shows how energy oscillates in the circuit, indicating the dynamic nature of power in AC systems.

Average Power

Average power in an LRC circuit represents the power consumed over an entire cycle. It highlights how continuous energy transfer happens despite instantaneous changes.
The formulas for average power across different components are:

• Resistor: The formula \(\langle p_R \rangle = \frac{1}{2} V_R I\) indicates that only resistors genuinely consume power as heat.
• Inductor and Capacitor: Their average power \(\langle p_L \rangle = 0\) and \(\langle p_C \rangle = 0\) imply no net energy consumption since they store and return energy to the circuit.

Understanding these values helps in designing efficient AC circuits, ensuring that energy is used where it's needed and not lost unnecessarily.

Sinusoidal Functions

Sinusoidal functions, intrinsic to AC circuits, are mathematical functions like sine and cosine that describe how variables such as voltage and current alternate over time.
These functions appear throughout LRC circuit analysis:

• The current functions as \(i = I \cos \omega t\), showing it's sinusoidal with time-dependent oscillations.
• Power expressions also depend on sinusoidal identities, such as \(\cos^2 \omega t = \frac{1}{2}(1 + \cos 2\omega t)\), simplifying power calculations.
• Sinusoidal behavior necessitates comprehension of phase relationships, frequencies, and amplitudes which affect how LRC circuits operate over time.

Understanding sinusoidal functions is crucial, as they form the foundation of analyzing AC circuit behavior and signal processing.

Power Conservation

Power conservation within an LRC circuit ensures that the energy input equals the sum of energies dissipated, stored, or returned by the components.
According to conservation principles:

• The power source delivers power characterized by \(p = V I \cos \omega t (\cos \phi \cos \omega t - \sin \phi \sin \omega t)\), where phase angles determine energy flow.
• Combining instantaneous powers \(p_R, p_L,\) and \(p_C\) should match the power from the source at any given time.
• This conservation implies that the power generated gets distributed among the components, maintaining equilibrium in the system.

Power conservation is fundamental in validating that a circuit operates correctly without losses or excess outside the system's control.

Phasor Analysis

Phasor analysis is an illustrative tool used to simplify AC circuit analysis by representing sinusoidal functions as vectors (phasors) in a complex plane.
For LRC circuits, phasors help:

• Visualize relationships between different elements' voltages and currents. Each component has a phasor illustrating its magnitude and phase relative to others.
• Analyze phase differences crucial for determining real power versus reactive power, aiding in resolving efficiency and power factor issues.
• Simplify calculations by transforming differential equations of time-dependent signals into algebraic equations, easing the solving process.

Phasor analysis facilitates a deeper understanding of how alternating currents and voltages interact, making it an essential part of AC circuit studies.