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Chapter 30: Problem 61

An \(L R\) circuit can be used as a "phase shifter." Assume that an "input" source voltage \(V=V_{0} \sin (2 \pi f t+\phi)\) is connected across a series combination of an inductor \(L=55 \mathrm{mH}\) and resistor \(R .\) The "output" of this circuit is taken across the resistor. If \(V_{0}=24 \mathrm{~V}\) and \(f=175 \mathrm{~Hz}\) determine the value of \(R\) so that the output voltage \(V_{R}\) lags the input voltage \(V\) by \(25^{\circ} .\) Compare (as a ratio) the peak output voltage with \(V_{0}\).

Short Answer

Expert verified
\( R \approx 129.72 \ \Omega \); Ratio: approximately 0.906.

Step by step solution

01

Understand the Phase Shift Equation

In an LR circuit, the voltage across the resistor lags the input voltage by an angle \( \theta \), which is determined by the formula: \( \tan(\theta) = \frac{X_L}{R} \), where \( X_L = 2 \pi f L \) is the inductive reactance. We need \( \theta = 25^\circ \).
02

Calculate the Inductive Reactance \(X_L\)

First, calculate the inductive reactance: \( X_L = 2 \pi f L = 2 \pi \times 175 \times 0.055 = 60.57 \ \Omega \).
03

Use the Phase Shift to Find \(R\)

We know \( \tan(25^\circ) = \frac{60.57}{R} \). Find \( R \) by rearranging the formula: \( R = \frac{60.57}{\tan(25^\circ)} \approx 129.72 \ \Omega \).
04

Determine the Ratio of Peak Output Voltage to \(V_0\)

The output voltage across the resistor \( V_R = V_0 \sin(\theta) \), hence the peak output voltage is \( V_{R_0} = V_0 \cdot \cos(\theta) \). Therefore, the peak output voltage is \( V_{R_0} = 24 \cdot \cos(25^\circ) \approx 21.74 \ \text{V}\). The ratio of the peak output voltage to the input voltage \( V_0 \) is \( \frac{21.74}{24} \approx 0.906 \).

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Key Concepts

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

Inductive Reactance
Inductive reactance is a core concept in analyzing LR circuits at AC frequencies. Inductors behave differently in AC circuits compared to DC circuits. In AC circuits, the voltage across the inductor isn't constant because the current is changing with time. This change leads to a phenomenon known as inductive reactance, symbolized as \(X_L\). Inductive reactance is given by the formula \(X_L = 2 \pi f L\), where \(f\) is the frequency of the AC source and \(L\) is the inductance of the coil.
  • Inductive reactance acts as a sort of resistance, but unlike regular resistance, it restricts the flow of the AC current based on frequency.
  • The higher the frequency or inductance, the higher the inductive reactance, meaning the circuit is more resistant to AC current flow.
A neat aspect of inductive reactance is how it causes the current to lag behind the voltage by 90 degrees in a purely inductive circuit. However, when mixed with resistors in an LR circuit, this phase relationship is modified.
Resistor Voltage Lag
In an LR circuit, the voltage across the resistor doesn't instantly follow the input voltage. Instead, it lags behind a bit. This lag is termed as the phase shift \(\theta\). It is influenced by the inductive reactance and resistance.
  • The relationship between these components is determined by \(\tan(\theta) = \frac{X_L}{R}\).
  • In our scenario, we have a desired phase shift of 25 degrees. This means that the resistor鈥檚 voltage is delayed by this angle in comparison to the input voltage.
Adjusting the resistance \(R\) lets us control the phase shift to favorably position the resistor voltage. This control is useful in applications requiring precise voltage timing, such as in oscillators and waveform generators. Understanding how the lag works helps in designing circuits that need specific phase relations between current and voltage.
Output Voltage Calculation
Calculating the output voltage in an LR circuit involves understanding the effect of the phase angle \(\theta\) on voltage magnitude. The output voltage \(V_R\) is derived from the input voltage and its sine component at the phase angle.
  • Given an input voltage expressed as \(V = V_0 \sin(2 \pi f t + \phi)\), the output voltage is \(V_R = V_0 \sin(\theta)\).
  • The peak output voltage \(V_{R_0}\) is thus calculated using \(V_{R_0} = V_0 \cdot \cos(\theta)\).
For instance, in the exercise, with an input peak voltage \(V_0\) of 24 V, and a phase angle of 25 degrees, the peak output voltage is 21.74 V. Comparing this peak voltage with the input voltage gives an idea of how much the voltage is delayed or reduced through the resistor. This ratio or fraction \(\frac{21.74}{24} \approx 0.906\) highlights the efficiency of voltage transfer from input to output, which is crucial in various electronic applications.

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Most popular questions from this chapter

(a) For an underdamped \(L R C\) circuit, determine a formula for the energy \(U = U _ { E } + U _ { B }\) stored in the electric and magnetic fields as a function of time. Give answer in terms of the initial charge \(Q _ { 0 }\) on the capacitor. (b) Show how \(d U / d t\) is related to the rate energy is transformed in the resistor, \(I ^ { 2 } R .\)

(a) Show that the self-inductance \(L\) of a toroid (Fig. 31\()\) of radius \(r _ { 0 }\) containing \(N\) loops each of diameter \(d\) is \(L \approx \frac { \mu _ { 0 } N ^ { 2 } d ^ { 2 } } { 8 r _ { 0 } }\) if \(r _ { 0 } \gg d .\) Assume the field is uniform inside the toroid; is this actually true? Is this result consistent with \(L\) for a solenoid? Should it be? (b) Calculate the inductance \(L\) of a large toroid if the diameter of the coils is 2.0\(\mathrm { cm }\) and the diameter of the whole ring is 66\(\mathrm { cm }\) . Assume the field inside the toroid is uniform. There are a total of 550 loops of wire.

(I) \(\mathrm { A } 75 - \Omega\) resistor and a \(6.8 - \mu \mathrm { F }\) capacitor are connected in series to an ac source. Calculate the impedance of the circuit if the source frequency is \(( a ) 60 \mathrm { Hz } ; ( b ) 6.0 \mathrm { MHz }\) .

(a) What is the rms current in an \(R C\) circuit if \(R=5.70 \mathrm{k} \Omega\), \(C=1.80 \mu \mathrm{F},\) and the rms applied voltage is \(120 \mathrm{~V}\) at \(60.0 \mathrm{~Hz} ?\) (b) What is the phase angle between voltage and current? (c) What is the power dissipated by the circuit? (d) What are the voltmeter readings across \(R\) and \(C\) ?

Determine the inductance \(L\) of the primary of a transformer whose input is 220\(\mathrm { V }\) at 60\(\mathrm { Hz }\) when the current drawn is 4.3\(\mathrm { A }\) . Assume no current in the secondary.
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