Skip to main content
Ch. 30 - Inductance, Electromagnetic Oscillations, and AC Circuits
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 29, Problem 88

Show that if the inductor L in the filter circuit of Fig. 30–33 (Problem 87) is replaced by a large resistor R, there will still be significant attenuation of the ac voltage and little attenuation of the dc voltage if the input dc voltage is high and the current (and power) are low.

Verified step by step guidance
1
Identify the key components of the circuit: The original circuit contains an inductor L, which is replaced by a resistor R. The goal is to analyze the behavior of the circuit for both AC and DC voltages.
Understand the behavior of a resistor in the circuit: A resistor R does not differentiate between AC and DC currents. However, the voltage drop across the resistor depends on the current flowing through it, as described by Ohm's Law: V=IR.
Analyze the DC voltage behavior: For a high input DC voltage, the current through the resistor will be low if the resistance R is large. This ensures that the power dissipation, given by P=IR2, remains low. Thus, the DC voltage is not significantly attenuated.
Analyze the AC voltage behavior: For AC signals, the resistor R will cause a voltage drop proportional to the current. If the resistance is large, the current will be small, leading to significant attenuation of the AC voltage. This is because the resistor opposes the rapid changes in current associated with AC signals.
Conclude the analysis: The circuit with a large resistor R will attenuate AC voltage significantly while allowing DC voltage to pass with little attenuation, provided the input DC voltage is high and the current (and power) are low. This behavior aligns with the desired filtering characteristics.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inductors in Filter Circuits

Inductors are passive components that store energy in a magnetic field when electrical current flows through them. In filter circuits, they are used to block high-frequency AC signals while allowing low-frequency DC signals to pass. The behavior of inductors is frequency-dependent, which is crucial for understanding how they affect voltage and current in circuits.
Recommended video:
Guided course
07:14
Inductors in AC Circuits

Resistors and Voltage Attenuation

Resistors are components that oppose the flow of electric current, leading to a drop in voltage across them. When a large resistor replaces an inductor in a filter circuit, it can still attenuate AC voltage significantly due to its resistance, while allowing DC voltage to remain relatively unaffected, especially when the input DC voltage is high and the current is low.
Recommended video:
Guided course
07:14
RMS Current and Voltage

AC vs. DC Voltage Behavior

AC (alternating current) and DC (direct current) behave differently in circuits. AC voltage varies with time, while DC voltage remains constant. In filter circuits, the design often aims to allow DC to pass through while blocking AC, which is essential for applications like power supplies. Understanding this distinction helps explain why replacing an inductor with a resistor can still achieve significant attenuation of AC voltage.
Recommended video:
Guided course
07:49
Alternating Voltages and Currents
Related Practice
Textbook Question

In some experiments, very tiny distances or spaces ( ≈ nm ) can be measured by using capacitance. Consider forming an LC circuit using a parallel-plate capacitor with plate area A, and a known inductance L. If f is on the order of 1 MHz and can be measured to a precision of ∆f = 1 Hz, with what percent accuracy can x be determined? Assume fringing effects at the capacitor’s edges can be neglected.

2
views
Textbook Question

In a plasma globe, a hollow glass sphere is filled with low-pressure gas and a small spherical metal electrode is located at its center. Assume an ac voltage source of peak voltage Vo and frequency f is applied between the metal sphere and the ground, and that a person is touching the outer surface of the globe with a fingertip, whose approximate area is 1.0 cm². The equivalent circuit for this situation is shown in Fig. 30–36, where RG and RP are the resistances of the gas and the person, respectively, and C is the capacitance formed by the gas, glass, and finger. (a) Determine C assuming it is a parallel-plate capacitor. The conductive gas and the person’s fingertip form the opposing plates of area A = 1.0 cm². The plates are separated by glass (dielectric constant K = 5.0) of thickness d = 2.0 mm. (b) In a typical plasma globe, f = 12 kHz. Determine the reactance XC of C at this frequency in MΩ. (c) The voltage may be Vo = 2500 V. With this high voltage, the dielectric strength of the gas is exceeded and the gas becomes ionized. In this “plasma” state, the gas emits light (“sparks”) and is highly conductive so that RG << XC. Assuming also that RP << XC, estimate the peak current that flows in the given circuit. Is this level of current dangerous? (d) If the plasma globe operated at f = 1.0 MHz, estimate the peak current that would flow in the given circuit. Is this level of current dangerous?


<IMAGE>

2
views
Textbook Question

In some experiments, very tiny distances or spaces ( ≈ nm ) can be measured by using capacitance. Consider forming an LC circuit using a parallel-plate capacitor with plate area A, and a known inductance L. When the plate separation is changed by ∆x, the circuit’s oscillation frequency will change by ∆f. Show that ∆x/x ≈ 2(∆f/f).

2
views
Textbook Question

For the circuit shown in Fig. 30–35, show that if the condition R₁ R₂ = L/C is satisfied then the potential difference between points a and b is zero for all frequencies.

2
views
Textbook Question

Filter circuit. Figure 30–33 shows a simple filter circuit designed to pass dc voltages with minimal attenuation and to remove, as much as possible, any ac components (such as 60-Hz line voltage that could cause hum in an audio system, for example). Assume Vin = V1 + V2 where V1 is dc and V2 = V20 sin ωt, and that any resistance is very small. (a) Determine the current through the capacitor: give amplitude and phase (assume R = 0 and XL > XC). (b) Show that the ac component of the output voltage, V2out, equals (Q/C) - V1 where Q is the charge on the capacitor at any instant, and determine the amplitude and phase of V2out (c) Show that the attenuation of the ac voltage is greatest when XC << XL, and calculate the ratio of the output to input ac voltage in this case. (d) Compare the dc output voltage to input voltage.

1
views
Textbook Question

Suppose a series LRC circuit has two resistors, R₁ and R₂, two capacitors, C₁ and C₂, and two inductors, L₁ and L₂ all in series. Calculate the total impedance of the circuit.

1
views