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Chapter 23: Problem 36

When a resistor is connected across the terminals of an ac generator \((112 \mathrm{~V})\) that has a fixed frequency, there is a current of \(0.500 \mathrm{~A}\) in the resistor. When an inductor is connected across the terminals of this same generator, there is a current of \(0.400 \mathrm{~A}\) in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what is (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator?

Short Answer

Expert verified
The impedance is approximately 358.4 惟, and the phase angle is 51.34掳.

Step by step solution

01

Calculate Resistance

First, find the resistance \( R \) using Ohm's law for the resistor: \( V = IR \).Substitute the given values: \[ 112 \mathrm{~V} = 0.500 \mathrm{~A} \times R \]Solve for \( R \):\[ R = \frac{112}{0.500} = 224 \; \Omega \]
02

Calculate Inductive Reactance

For the inductor, use the formula \( V = I L\omega \) for inductive reactance, here \( V = I X_L \), where \( X_L \) is the reactance.Substitute the given values: \[ 112 \mathrm{~V} = 0.400 \mathrm{~A} \times X_L \]Solve for \( X_L \):\[ X_L = \frac{112}{0.400} = 280 \; \Omega \]
03

Calculate Impedance

For the series circuit, impedance \( Z \) can be calculated using the formula \[ Z = \sqrt{R^2 + X_L^2} \].Substitute the calculated values of \( R \) and \( X_L \):\[ Z = \sqrt{224^2 + 280^2} \]\[ Z = \sqrt{50176 + 78400} \]\[ Z = \sqrt{128576} \approx 358.4 \; \Omega \]
04

Calculate Phase Angle

The phase angle \( \phi \) between the current and voltage is given by the formula \[ \tan \phi = \frac{X_L}{R} \].Substitute the values for \( X_L \) and \( R \):\[ \tan \phi = \frac{280}{224} \]\[ \tan \phi = 1.25 \]Calculate \( \phi \) using the tangent inverse function:\[ \phi = \arctan(1.25) \approx 51.34^\circ \]

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Key Concepts

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

Ohm's Law
One of the most fundamental concepts in AC circuits is Ohm's Law. It relates the voltage (V), current (I), and resistance (R) in a circuit. In formula form, it states that \( V = IR \). This relationship shows how the voltage supplied by a generator is distributed across a resistor, affecting the flow of current.
In our example, with a resistor connected to a 112 V AC generator and a current of 0.500 A, we use Ohm's Law to find the resistance \((R)\). The calculation goes like this:
  • Plug in the values to get \( 112 = 0.500 \times R \).
  • Solving for \( R \) gives \( R = \frac{112}{0.500} = 224 \; \Omega \).
This calculation shows how the electrical resistance in a circuit dictates how much current will flow for a given voltage.
Impedance
Impedance is a key concept in AC circuits that extends the idea of resistance to include the effects of circuit elements like capacitors and inductors. While resistance applies to direct current (DC) circuits, impedance takes into account the frequency of the alternating current.
In AC circuits with components in series, the total impedance \( Z \) is calculated with the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( R \) is resistance and \( X_L \) is inductive reactance.
  • In our problem, first, find the resistance \( R = 224 \Omega \).
  • Then, determine inductive reactance as \( X_L = 280 \Omega \).
  • Calculate total impedance: \( Z = \sqrt{224^2 + 280^2} \approx 358.4 \Omega \).
Impedance gives us a broad understanding of how the circuit opposes the flow of AC current.
Inductive Reactance
Inductive reactance is the resistance the circuit offers to the flow of AC current due to the presence of inductors. It is represented by \( X_L \) and depends on the frequency of the current and the inductance of the coil. The relationship is defined by \( X_L = L \omega \), where \( \omega \) is the angular frequency.
In the exercise, we use a shortcut: \( V = I X_L \), where \( V \) is the voltage and \( I \) is the current:
  • Given \( V = 112 \) V and \( I = 0.400 \) A for the inductor.
  • Calculate \( X_L = \frac{112}{0.400} = 280 \; \Omega \).
The inductive reactance signifies how much the inductor will resist changes in the current within the circuit.
Phase Angle
The phase angle \( \phi \) is an essential aspect of AC circuits. It shows the phase difference between the voltage across the circuit and the current flowing through it. This phase difference arises because of the reactance components that shift the current with respect to the voltage.
The phase angle is calculated by the formula:
\[ \tan \phi = \frac{X_L}{R} \]
  • For the problem, with \( X_L = 280 \; \Omega \) and \( R = 224 \; \Omega \).
  • Find \( \tan \phi = \frac{280}{224} = 1.25 \).
  • Use the inverse tangent to find \( \phi = \arctan(1.25) \approx 51.34^\circ \).
The phase angle tells us how "out of sync" the current is with the voltage, which is crucial for understanding the behavior of AC circuits.

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Most popular questions from this chapter

Concept Questions Part \(a\) of the drawing shows a resistor and a charged capacitor wired in series. When the switch is closed, the capacitor discharges as charge moves from one plate to the other. Part \(b\) shows a plot of the amount of charge remaining on each plate of the capacitor as a function of time. (a) What does the time constant \(\tau\) of this resistor-capacitor circuit physically represent? (b) How is the time constant related to the resistance \(R\) and the capacitance \(C ?\) (c) In part \(c\) of the drawing, the switch has been removed and an ac generator has been inserted into the circuit. What is the impedance \(Z\) of this circuit? Express your answer in terms of the resistance \(R,\) the time constant \(\tau,\) and the frequency \(f\) of the generator. Problem The circuit elements in the drawing have the following values: \(R=18 \Omega\) \(V_{\mathrm{rms}}=24 \mathrm{~V}\) for the generator, and \(f=380 \mathrm{~Hz}\). The time constant for the circuit is \(\tau=3.0 \times 10^{-4} \mathrm{~s} .\) What is the rms current in the circuit?

A capacitor (capacitance \(C_{1}\) ) is connected across the terminals of an ac generator. Without changing the voltage or frequency of the generator, a second capacitor (capacitance \(C_{2}\) ) is added in series with the first one. As a result, the current delivered by the generator decreases by a factor of three. Suppose the second capacitor had been added in parallel with the first one, instead of in series. By what factor would the current delivered by the generator have increased?

A capacitor is connected across an ac generator whose frequency is \(750 \mathrm{~Hz}\) and whose peak output voltage is \(140 \mathrm{~V}\). The rms current in the circuit is \(3.0 \mathrm{~A}\). (a) What is the capacitance of the capacitor? (b) What is the magnitude of the maximum charge on one plate of the capacitor?

A \(2700-\Omega\) resistor and a \(1.1-\mu F\) capacitor are connected in series across a generator \((60.0 \mathrm{~Hz}, 120 \mathrm{~V}) .\) Determine the power delivered to the circuit.

A series RCL circuit has a resonant frequency of \(690 \mathrm{kHz}\). If the value of the capacitance is \(2.0 \times 10^{-9} \mathrm{~F}\), what is the value of the inductance?
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