/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 An \(80.0-\Omega\) resistor and ... [FREE SOLUTION] | 91Ó°ÊÓ

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An \(80.0-\Omega\) resistor and a \(200-\mathrm{mH}\) inductor are connected in parallel across a \(100-\mathrm{V}\) (rms), \(60.0-\mathrm{Hz}\) source. (a) What is the rms current in the resistor? (b) By what angle does the total current lead or lag behind the voltage?

Short Answer

Expert verified
The RMS current in the resistor is calculated in Step 2, and the phase difference between the total current and voltage is calculated in Step 5. If necessary perform the arithmetic operation accordingly to obtain the precise results.

Step by step solution

01

Calculate the resistance and reactance

Resistance \( R \) is given as 80.0 Ω, and reactance \( X_L = 2\pi f L \), where \( f = 60.0 Hz \) and \( L = 200mH = 200 x 10^{-3} H \) . Let's proceed to calculate \( X_L \).
02

Find the rms current in the resistor

The rms current \( I_R \) in the resistor is given by \( I_R = V/R \), where \( V = 100V \) and \( R = 80.0 \Omega \). Product the calculation to find \( I_R \).
03

Calculate the current in the inductor

The rms current \( I_L \) in the inductor is given by \( I_L = V/X_L \). Use the values for \( V \) and \( X_L \) calculated in step one to find \( I_L \).
04

Find the total rms current

The total rms current \( I \) is given by \( I = sqrt(I_R^2 + I_L^2 ) \). Use the calculated values for \( I_R \) and \( I_L \) to find \( I \).
05

Calculate the phase difference

The phase difference \( \theta \) is given by \( \theta = arctan(I_L/I_R \). Use the values for \( I_L \) and \( I_R \) to calculate \( \theta \). The current is said to lead the voltage if \( \theta > 0 \) and lag if \( \theta < 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistor Inductance
In AC circuits, inductance and resistance are often encountered together, especially in parallel and series combinations. Inductance is a property of an inductor that opposes changes in current. When an AC voltage is applied, the inductor builds up a magnetic field, which resists the changes in current that AC inherently presents.
To understand resistor-inductance setups, we need to grasp a bit of physics.- **Resistance** (R) signifies the opposition to the flow of direct current, measured in ohms (Ω).- **Inductance** (L), given in henrys (H), is a measure of a circuit's ability to induce voltage as a response to changes in current.- **Reactance** (X_L), a type of "resistance" to changing current, depends on the frequency of the AC source and is calculated with the formula: \( X_L = 2\pi f L \) Where \( f \) is the frequency of the AC source. In practical terms, this means that as frequency increases, the reactance of an inductor in an AC circuit increases, making it more difficult for the current to pass through. This is crucial when analyzing how voltage and current relate in circuits containing both resistors and inductors.
RMS Current
RMS, or root mean square, is a method used to express AC values as if they were direct current (DC) values. This concept is particularly useful in AC circuits, like the one in our exercise. It allows us to compare AC power to DC power.
To find the **RMS current** in a circuit with resistive and inductive components, there are a few steps:- For a resistor, the RMS current \( I_R \) is simply: \( I_R = \frac{V}{R} \) Here, \( V \) is the RMS voltage across the resistor, and \( R \) is its resistance.- For an inductor, the RMS current \( I_L \) is calculated using the reactance \( X_L \): \( I_L = \frac{V}{X_L} \) - The total RMS current \( I \) in a circuit with just parallel inductive and resistive elements is found by calculating the square root of the sum of the squares of \( I_R \) and \( I_L \): \[ I = \sqrt{I_R^2 + I_L^2} \] This shows how much actual current is "delivered" in the effective combination of both resistor and inductive properties in AC circuits, giving students foundational concepts they need to solve AC circuit problems.
Phase Angle
Understanding the phase angle is key when studying AC circuits, as it tells us how far out of sync the voltage and the current are with each other. A phase angle arises due to differing properties of resistors and inductors.- **Definition**: The phase angle \( \theta \) in a circuit indicates the phase difference between voltage and current. Calculated as: \( \theta = \text{arctan}\left(\frac{I_L}{I_R}\right) \) Here, \( I_L \) is the RMS current due to the inductor, while \( I_R \) is due to the resistor.- **Interpretation**: - If \( \theta > 0 \), the current leads the voltage (common in capacitive circuits). - If \( \theta < 0 \), the current lags behind the voltage (typical in inductive circuits).This phase angle is critical because it affects how the circuit's power is delivered and distributed. It can determine the efficiency of AC power delivery, with different components contributing uniquely to the total circuit behavior. A solid grasp of this concept ensures better understanding and optimization of AC circuits.

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Most popular questions from this chapter

A series \(R L C\) circuit has components with following values: \(L=20.0 \mathrm{mH}, C=100 \mathrm{nF}, R=20.0 \Omega,\) and \(\Delta V_{\max }=100 \mathrm{V}\) with \(\Delta v=\Delta V_{\ln \ln x} \sin \omega t .\) Find \((a)\) the resonant frequency, (b) the amplitude of the current at the resonant frequency, (c) the \(Q\) of the circuit, and (d) the amplitude of the voltage across the inductor at resonance.

A resistor \(R\), inductor \(L\), and capacitor \(C\) are connected in series to an AC source of rms voltage \(\Delta V\) and variable frequency. Find the energy that is delivered to the circuit during one period if the operating frequency is twice the resonance frequency.

The rms output voltage of an AC source is \(200 \mathrm{V}\) and the operating frequency is \(100 \mathrm{Hz}\). Write the equation giving the output voltage as a function of time.

A step-down transformer is used for recharging the batteries of portable devices such as tape players. The turns ratio inside the transformer is 13: 1 and it is used with \(120-\mathrm{V}\) (rms) household service. If a particular ideal transformer draws \(0.350 \mathrm{A}\) from the house outlet, what are (a) the voltage and (b) the current supplied to a tape player from the transformer? (c) How much power is delivered?

An inductor \((L=400 \mathrm{mH}),\) a capacitor \((C=4.43 \mu \mathrm{F})\) and a resistor \((R=500 \Omega)\) are connected in series. A \(50.0-\mathrm{Hz}\) AC source produces a peak current of \(250 \mathrm{mA}\) in the circuit. (a) Calculate the required peak voltage \(\Delta V_{\text {max: }}\) (b) Determine the phase angle by which the current leads or lags the applied voltage.

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