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Two coils wound on a common core have \(L_{1}=1 \mathrm{H}, L_{2}=2 \mathrm{H}\), and \(M=0.5 \mathrm{H}\). The currents are \(i_{1}=\cos (377 t) \mathrm{A}\) and \(i_{2}=\) \(0.5 \cos (377 t)\) A. Both of the currents enter dotted terminals. Find expressions for the voltages across the coils.

Short Answer

Expert verified
The voltage across coil 1 is \(-471.25 \sin(377t)\) V, and across coil 2 is \(-565.5 \sin(377t)\) V.

Step by step solution

01

Understand the Induced Voltage Formula

The voltage across an inductor is given by the formula: \(v = L \cdot \frac{di}{dt}\). In the case of mutual inductance, the voltage across one coil is affected by the change in current in the other coil as well. The voltage for coil 1 and coil 2 are given by \( v_1 = L_1 \cdot \frac{di_1}{dt} + M \cdot \frac{di_2}{dt} \) and \( v_2 = L_2 \cdot \frac{di_2}{dt} + M \cdot \frac{di_1}{dt} \), respectively.
02

Derive Current Derivatives

Calculate the derivatives of the current functions. For \(i_1 = \cos(377t)\), \(\frac{di_1}{dt} = -377 \sin(377t)\). For \(i_2 = 0.5 \cos(377t)\), \(\frac{di_2}{dt} = -188.5 \sin(377t)\).
03

Calculate Voltage Across Coil 1

Substitute the values and derivatives into the formula for \(v_1\):\[ v_1 = L_1 \cdot \frac{di_1}{dt} + M \cdot \frac{di_2}{dt} = 1 \cdot (-377 \sin(377t)) + 0.5 \cdot (-188.5 \sin(377t)) \]This simplifies to:\[ v_1 = (-377 - 94.25) \sin(377t) \]\[ v_1 = -471.25 \sin(377t) \].
04

Calculate Voltage Across Coil 2

Substitute the values and derivatives into the formula for \(v_2\):\[ v_2 = L_2 \cdot \frac{di_2}{dt} + M \cdot \frac{di_1}{dt} = 2 \cdot (-188.5 \sin(377t)) + 0.5 \cdot (-377 \sin(377t)) \]This simplifies to:\[ v_2 = (-377 - 188.5) \sin(377t) \]\[ v_2 = -565.5 \sin(377t) \].

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Key Concepts

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

Induced Voltage
Induced voltage is a crucial concept in understanding how electrical circuits with inductors operate. When we talk about induced voltage, we are referring to the potential difference that is created across an inductor due to its interaction with changing current. In essence, inductors resist changes in the current flowing through them. Thus, when the current changes, whether by increasing or decreasing, an electromotive force (emf) is induced. This is known as induced voltage.

The key to finding the voltage in our exercise is using the formula for the voltage across an inductor:
  • The voltage across a single inductor is given by: \( v = L \cdot \frac{di}{dt} \), where \( L \) is inductance and \( \frac{di}{dt} \) is the rate of change of current with respect to time.
  • In circuits with mutual inductance, the voltage induced in one coil also depends on the rate of change of current in the other coil. This is represented by: \( v_1 = L_1 \cdot \frac{di_1}{dt} + M \cdot \frac{di_2}{dt} \) and \( v_2 = L_2 \cdot \frac{di_2}{dt} + M \cdot \frac{di_1}{dt} \), where \( M \) is mutual inductance.
Understanding induced voltage helps to predict the behavior of circuits in dynamics and is fundamental in designing transformers and other electromechanical devices.
Coil Inductance
Coil inductance is an intrinsic property of an inductor that quantifies its ability to resist changes in current. It depends on factors like the number of wire loops (turns) in the coil, the core material, and the coil's dimensions. Inductance is measured in henries (H).

When dealing with multiple coils, we also need to consider mutual inductance, especially if they are closely positioned like in our exercise:
  • **Self-Inductance (L):** It is the property of a coil that resists the change in current flowing through it. Every inductor has its own self-inductance defined by \( L_1 \), \( L_2 \), etc.
  • **Mutual Inductance (M):** This occurs when two inductors are placed such that the magnetic field of one affects the other, like the exercise where two coils affect each other's output. The changing magnetic field in one coil induces a voltage in the nearby coil, and vice versa. The mutual inductance is determined by layout and the closeness of the coils.
In practical applications, understanding mutual inductance is vital for the design and function of transformers and inductive coupling circuits, as it can either enhance or limit their performance depending on configuration.
Current Derivatives
The rate at which current changes over time is crucial in understanding how voltages are induced in inductors. This rate is expressed mathematically as a derivative, noted as \( \frac{di}{dt} \). In our exercise, the precise calculation of these derivatives is key in determining the induced voltages across the coils.

Taking derivatives means calculating the slope of the function at a given point, indicating how rapidly the current is changing:
  • For a cosine function like \( i_1 = \cos(377t) \), the derivative is \( \frac{di_1}{dt} = -377 \sin(377t) \).
  • Similarly, for \( i_2 = 0.5 \cos(377t) \), its derivative is \( \frac{di_2}{dt} = -188.5 \sin(377t) \).
Such derivatives help understand how quickly the magnetic field that surrounds an inductor change, determining how much and how quickly voltage will be induced. Mathematically, they represent the core of dynamic circuit analysis, translating real-world current fluctuations into impactful voltage changes.

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Most popular questions from this chapter

A \(60-\mathrm{Hz} 40-\mathrm{kVA} 8000 / 240-\mathrm{V}-\mathrm{rms}\) transformer has the following equivalent-circuit parameters: $$ \begin{array}{lcl} \hline \text { Primary resistance } & R_{1} & 20 \Omega \\ \text { Secondary resistance } & R_{2} & 0.04 \Omega \\ \text { Primary leakage reactance } & X_{1}=\omega L_{1} & 100 \Omega \\ \text { Secondary leakage reactance } & X_{2}=\omega L_{2} & 0.2 \Omega \\ \text { Magnetizing reactance } & X_{m}=\omega L_{m} & 40 \mathrm{k} \Omega \\\ \text { Core-loss resistance } & R_{c} & 200 \mathrm{k} \Omega \\ \hline \end{array} $$ Find the percentage regulation and power efficiency for the transformer for a rated load having a lagging power factor of \(0.9\).

A transformer is needed that will cause an actual load resistance of \(25 \Omega\) to appear as \(100 \Omega\) to an ac voltage source of \(240 \mathrm{~V} \mathrm{rms}\). Draw the diagram of the circuit required. What turns ratio is required for the transformer? Find the current taken from the source, the current flowing through the load, and the load voltage.

Two coils wound on a common core have \(L_{1}=0.2 \mathrm{H}, L_{2}=0.5 \mathrm{H}\), and \(M=0.1 \mathrm{H}\). The currents are \(i_{1}=\exp (-1000 t) \mathrm{A}\) and \(i_{2}=2 \exp (-1000 t)\) A. Both of the currents enter dotted terminals. Find expressions for the voltages across the coils.

A uniform flux density given by \(B=0.5 \sin\) \((377 t) \mathrm{T}\) is perpendicular to the plane of a 1000 -turn circular coil of radius \(10 \mathrm{~cm}\). Find the flux linkages and the voltage as functions of time. Find their maximum values also.

A 1000-turn coil is wound on an iron core. When a \(120-\mathrm{V}-\mathrm{rms} 60-\mathrm{Hz}\) voltage is applied to the coil, the current is 2 A rms. Neglect the resistance of the coil. Determine the reluctance of the core. Given that the crosssectional area of the core is \(5 \mathrm{~cm}^{2}\) and the length is \(20 \mathrm{~cm}\), determine the relative permeability of the core material.

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