/*! 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 95 Determine the inductance \(L\) o... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The inductance \( L \) of the primary coil is approximately 0.1358 H.

Step by step solution

01

Understand the Formula

The inductance of a transformer primary coil can be found using the formula for inductive reactance \( X_L \), which is given by \( X_L = 2 \pi f L \), where \( f \) is the frequency and \( L \) is the inductance. Inductive reactance is also given by \( X_L = \frac{V}{I} \), where \( V \) is the voltage and \( I \) is the current.
02

Calculate Inductive Reactance

Calculate the inductive reactance \( X_L \) using \( X_L = \frac{V}{I} \) with \( V = 220 \ \mathrm{V} \) and \( I = 4.3 \ \mathrm{A} \). \[ X_L = \frac{220}{4.3} \approx 51.16 \ \Omega \] This is the measure of the reactance in ohms.
03

Solve for Inductance

Using the formula \( X_L = 2 \pi f L \), solve for \( L \). Plug in \( X_L = 51.16 \ \Omega \) and \( f = 60 \ \mathrm{Hz} \). Rearrange to find \( L \): \[ L = \frac{X_L}{2 \pi f} = \frac{51.16}{2 \pi \times 60} \approx 0.1358 \ \mathrm{H} \] Calculating this gives the inductance in henries.

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Key Concepts

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

Transformer
A transformer is an electrical device that transfers electrical energy between two or more circuits through electromagnetic induction. It is essential in adjusting voltage levels, either stepping them up or stepping them down, to make the power distribution more efficient.
Transformers consist of two main coils: the primary and the secondary. The primary coil is where input voltage is applied, and the secondary coil is where the output voltage is obtained. The voltage change between these coils is determined by the ratio of the number of turns in the primary coil to the number of turns in the secondary coil.
- **Key Points**: - Transform efficiency supports long-distance power distribution. - Ideally, power input in equals power output, minus losses. - Transformers do not change power frequency; they merely alter voltage levels.
This example focuses on the primary side of the transformer, where inductance is vital to its operation. The inductance of the primary winding affects how the transformer responds to an alternating current (AC) input signal.
Inductive Reactance
Inductive reactance is an important concept in AC circuits, especially when dealing with components like transformers that include coils or inductors. It arises because inductors resist changes in current. This resistance creates a sort of impeding force against the AC, which we calculate as reactance, noted as \( X_L \). - **Formula Recap**: - \( X_L = 2 \pi f L \) - \( X_L = \frac{V}{I} \)
In transformers, the inductive reactance relates to how the primary coil resists the AC current due to its inductance. From impedance point of view, as frequency increases, so does reactance, impacting how much AC current passes through the transformer.
In our exercise, you use these concepts to find the inductance by knowing the voltage, current, and frequency. Inductive reactance helps predict how the inductor behaves in an AC environment, making it crucial in designing circuits that include inductors or transformers.
AC Circuit Analysis
Analyzing AC circuits involves understanding how different components, such as resistors, capacitors, and inductors, react to AC signals. These reactions are important for correctly predicting the behavior of electrical devices like transformers in an AC setting.
When we perform AC circuit analysis, we consider the complex nature of AC signals: - **Amplitude**: Represents how strong the signal is. - **Frequency**: Describes how often the signal cycles per second. - **Phase**: Indicates how far along its cycle a wave is at any given time.
Understanding how inductance impacts an AC circuit is key. Inductors resist changes to current, thus affecting amplitude and phase relationships in the circuit. When performing AC circuit analysis, you predict these effects using parameters like inductance and reactance.
For example, to determine the inductance in a transformer's primary coil, analyze the AC performance with given frequency, voltage, and current values. This approach helps better manage and design electrical systems that involve varying AC inputs.

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Most popular questions from this chapter

(II) In some experiments, short distances are measured by using capacitance. Consider forming an \(L C\) circuit using a parallel-plate capacitor with plate area \(A ,\) and a known inductance \(L . ( a )\) If charge is found to oscillate in this circuit at frequency \(f = \omega / 2 \pi\) when the capacitor plates are separated by distance \(x ,\) show that \(x = 4 \pi ^ { 2 } A \epsilon _ { 0 } f ^ { 2 } L\) . (b) When the plate separation is changed by \(\Delta x ,\) the circuit's oscillation frequency will change by \(\Delta f .\) Show that \(\Delta x / x \approx 2 ( \Delta f / f ) . ( c )\) If \(f\) is on the order of 1\(\mathrm { MHz }\) and can be measured to a precision of \(\Delta f = 1 \mathrm { Hz } ,\) with what percent accuracy can \(x\) be determined? Assume fringing effects at the capacitor's edges can be neglected.

To detect vehicles at traffic lights, wire loops with dimensions on the order of 2\(\mathrm { m }\) are often buried horizontally under roadways. Assume the self-inductance of such a loop is \(L = 5.0 \mathrm { mH }\) and that it is part of an \(L R C\) circuit as shown in Fig. 37 with \(C = 0.10 \mu \mathrm { F }\) and \(R = 45 \Omega\) . The ac voltage has frequency \(f\) and rms voltage \(V _ { \mathrm { rms } }\) . (a) The frequency \(f\) is chosen to match the resonant frequency \(f _ { 0 }\) of the circuit. Find \(f _ { 0 }\) and determine what the rms voltage \(\left( V _ { R } \right) _ { \text { Tms across the resistor will be when } }\) \(f = f _ { 0 } . ( b )\) Assume that \(f , C ,\) and \(R\) never change, but that, when a car is located above the buried loop, the loop's self-inductance decreases by 10\(\%\) (due to induced eddy currents in the car's metal parts). Determine by what factor the voltage \(\left( V _ { R } \right) _ { \text { rms decreases in this situation in } }\) comparison to no car above the loop. [Monitoring \(\left( V _ { R } \right) _ { \mathrm { rms } }\) detects the presence of a car.

(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) What is the rms current in a series \(L R\) circuit when a \(60.0-\mathrm{Hz}, 120-\mathrm{V}\) rms ac voltage is applied, where \(R=965 \Omega\) and \(L=225 \mathrm{mH} ? \quad\) (b) What is the phase angle between voltage and current? (c) How much power is dissipated? (d) What are the rms voltage readings across \(R\) and \(L ?\)

(I) The variable capacitor in the tuner of an AM radio has a capacitance of \(1350 \mathrm{pF}\) when the radio is tuned to a station at \(550 \mathrm{kHz}\). ( \(a\) ) What must be the capacitance for a station at \(1600 \mathrm{kHz} ?\) (b) What is the inductance (assumed constant)? Ignore resistance.

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