Problem 20 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit, the components have the following values: \(L=20.0 \mathrm{mH}, C=140 \mathrm{nF},\) and \(R=350 \Omega .\) The generator has an rms voltage of 120 \(\mathrm{V}\) and a frequency of 1.25 \(\mathrm{kHz}\) . Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

Step-by-Step Solution

Answer

The power supplied by the generator is approximately 4.65 W, and the power dissipated in the resistor is approximately 4.69 W.

Step 1: Calculate Angular Frequency

First, we need to find the angular frequency, \(\omega\), of the circuit using the formula \(\omega = 2\pi f\), where \(f\) is the frequency of the generator. Given \(f = 1.25 \text{ kHz}\), we have:\[\omega = 2\pi \times 1250 \text{ Hz} = 2500\pi \text{ rad/s}.\]

Step 2: Calculate Inductive Reactance

Use the inductance \(L\) to find the inductive reactance, \(X_L\), using the formula \(X_L = \omega L\). With \(L = 20 \text{ mH} = 20 \times 10^{-3} \text{ H}\), we calculate:\[X_L = 2500\pi \times 0.020 = 50\pi \; \Omega.\]

Step 3: Calculate Capacitive Reactance

Find the capacitive reactance, \(X_C\), using the formula \(X_C = \frac{1}{\omega C}\). Given \(C = 140 \text{ nF} = 140 \times 10^{-9} \text{ F}\), we have:\[X_C = \frac{1}{2500\pi \times 140 \times 10^{-9}} \approx 1.14 \times 10^3 \; \Omega.\]

Step 4: Calculate Impedance

Determine the total impedance, \(Z\), using the formula \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). Substituting the known values:\[Z = \sqrt{350^2 + (50\pi - 1140)^2} \approx \sqrt{350^2 + (-982.15)^2} \approx 1035 \, \Omega.\]

Step 5: Calculate RMS Current

The rms current, \(I_{rms}\), is calculated using Ohm's Law: \(I_{rms} = \frac{V_{rms}}{Z}\), where \(V_{rms} = 120\, \text{V}\). Thus:\[I_{rms} = \frac{120}{1035} \approx 0.116 \text{ A}.\]

Step 6: Calculate Power Supplied by the Generator

The average power supplied by the generator is given by \(P = V_{rms} \times I_{rms} \times \cos\phi\), where \(\cos\phi = \frac{R}{Z}\). We calculate \(\cos\phi\) first:\[\cos\phi = \frac{350}{1035} \approx 0.338.\]Thus, the power is:\[P = 120 \times 0.116 \times 0.338 \approx 4.65 \text{ W}.\]

Step 7: Calculate Power Dissipated in the Resistor

The power dissipated in the resistor, \(P_R\), is given by \(P_R = I_{rms}^2 \times R\). Thus:\[P_R = (0.116)^2 \times 350 \approx 4.69 \text{ W}.\]

Key Concepts

Reactance Calculation

In an L-R-C circuit, reactance is a critical component when analyzing the behavior of inductors and capacitors in response to an alternating current (AC). Inductive reactance, denoted as \(X_L\), is determined by how an inductor resists changes in current and is calculated using the formula \(X_L = \omega L\), where \(\omega\) is the angular frequency and \(L\) is the inductance. For capacitors, the capacitive reactance, denoted as \(X_C\), reflects their ability to block changes in voltage, calculated by \(X_C = \frac{1}{\omega C}\), with \(C\) being the capacitance. Each type of reactance has an opposite effect on the circuit’s overall phase angle and impacts the impedance. These calculations are important for determining how energy is stored within the circuit's electromagnetic and electric fields.

Impedance in AC Circuits

Impedance, symbolized by \(Z\), is the total resistance to the flow of alternating current, combining both resistive (real) and reactive (imaginary) effects. It’s more comprehensive than simple resistance encountered in direct current circuits. The impedance in an L-R-C circuit is calculated using the formula \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(R\) is the resistance, and \(X_L\) and \(X_C\) are the inductive and capacitive reactances respectively. This equation takes into account the phase angle differences between the voltage and current due to reactance. Understanding impedance is crucial for analyzing power flow and voltage/current relationships in AC powered systems. A complex impedance influences how efficiently power is transmitted and can affect the performance of electrical devices connected to the circuit.

Power in AC Circuits

Power in AC circuits differs from that in DC circuits due to the phase difference between current and voltage, which impacts how power is calculated and interpreted. The true or real power \(P\), given in watts, represents the power actually used in the circuit. It's computed as \(P = V_{rms} \times I_{rms} \times \cos\phi\), where \(V_{rms}\) is the root mean square voltage, \(I_{rms}\) is the root mean square current, and \(\cos\phi\) (power factor) reflects the cosine of the phase angle \(\phi\).

• The power factor indicates the efficiency of power usage.
• A power factor of 1 means all power is used for work; lower values indicate more power is wasted as reactive power.

In an L-R-C circuit, the power dissipated by the resistor, \(P_R = I_{rms}^2 \times R\), is always part of the real power since resistors consume energy. The remaining power, which oscillates back and forth in the circuit, is stored in the reactive components. Understanding these distinctions allows for improved design and operation of electrical systems.