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Chapter 31: Problem 66

31.66. A transformer consists of 275 primary windings and 834 secondary windings. If the potential difference across the primary coil is \(25.0 \mathrm{V},\) (a) what is the voltage across the secondary coil, and (b) what is the effective load resistance nf the secondary coil if it is connected across a \(125-\Omega\) resistor?

Short Answer

Expert verified
(a) 75.82 V, (b) 13.7 惟

Step by step solution

01

Understanding the Transformer Relation

The transformer is based on the principle that the ratio of the primary to secondary voltage is equal to the ratio of the number of turns in the primary and secondary coils. This can be stated as \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), where \( V_s \) is the secondary voltage, \( V_p \) is the primary voltage, \( N_s \) is the number of secondary windings, and \( N_p \) is the number of primary windings.
02

Calculating Secondary Voltage

Given that \( V_p = 25.0 \mathrm{V} \), \( N_p = 275 \), and \( N_s = 834 \), substitute these values into the transformer equation: \( \frac{V_s}{25.0} = \frac{834}{275} \). Solve for \( V_s \) to find that \( V_s = \frac{834}{275} \times 25.0 \approx 75.82 \mathrm{V} \).
03

Calculating Effective Load Resistance

Since power is conserved in an ideal transformer, \( V_p I_p = V_s I_s \). The effective load resistance \( R_{eff} \) can be calculated as \( R_{eff} = \left(\frac{V_p}{V_s}\right)^2 \times R_L \), where \( R_L = 125 \Omega \) is the load resistance. Substitute \( V_p = 25.0 \mathrm{V} \) and \( V_s = 75.82 \mathrm{V} \) to find \( R_{eff} = \left(\frac{25.0}{75.82}\right)^2 \times 125 \Omega \approx 13.7 \Omega \).

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Key Concepts

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

Voltage Transformation Ratio
In transformers, the voltage transformation ratio is a vital concept that directly links the number of windings in the coils to the transformation of voltage levels.
This ratio can be expressed with the equation:
  • \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
  • where \( V_s \) and \( V_p \) are the secondary and primary voltages, respectively.
  • \( N_s \) and \( N_p \) are the number of turns in the secondary and primary windings, respectively.

The equation reveals that:
  • When the number of turns in the secondary coil \( N_s \) is greater than in the primary \( N_p \), the voltage increases鈥攁 step-up transformer.
  • If the secondary turns are fewer, it's a step-down transformer.

By understanding this ratio, one can easily determine how a transformer's construction (by adjusting the number of coil windings) affects the voltage outcome, essential for applications requiring varied voltage levels.
Effective Load Resistance
The concept of effective load resistance is crucial in understanding how transformers influence the load they are connected to, from the perspective of the primary coil.
In an ideal transformer, the formula for effective load resistance, \( R_{eff} \), is given by:
  • \( R_{eff} = \left(\frac{V_p}{V_s}\right)^2 \times R_L \)
  • where \( V_p \) and \( V_s \) are the primary and secondary voltages respectively.
  • \( R_L \) is the load resistance connected to the secondary winding.
This formula can help us understand:
  • How the secondary resistor appears when viewed from the primary side in terms of its resistance.
  • The impact of changing voltage levels via the transformer on perceived resistance.
Thus, modifying the effective load resistance can manage how much current the primary coil needs to supply, aligning electrical load management with efficiency goals.
Ideal Transformer Assumptions
In analyzing transformers, especially in theoretical exercises, it's common to employ the ideal transformer assumptions. These assumptions simplify calculations and serve as a basic model for understanding transformer behavior.

Here are the main assumptions made for an ideal transformer:
  • No power losses: \( V_p I_p = V_s I_s \), implying that all power input is transferred without loss.
  • No resistance in windings: both the primary and secondary windings' resistance is negligible.
  • High permeability core: ensures that magnetic coupling between the coils is perfect, with no flux leakage.
  • Infinite open-circuit inductance: this means that there is no magnetizing current unless necessary for perfect operation.
  • 100% efficiency: assumes all energy put into one side can be fully extracted from the other.

While these assumptions don't hold in real-life transformers (due to inefficiencies like core losses and winding resistances), they provide a strong foundation for understanding core transformer principles and designing real systems that approach these ideals.

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Most popular questions from this chapter

31.51. An \(L-R-C\) series circuit is connected to an ac source nf constant voltage amplitude \(V\) and variable angular frequency \(\omega\) . (a) Show that the current amplitude, as a function of \(\omega,\) is $$ I=\frac{V}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}} $$ (b) Show that the average power dissipated in the resistor is $$ P=\frac{V^{2} R / 2}{R^{2}+(\omega L-1 / \omega C)^{2}} $$ (c) Show that \(I\) and \(P\) are both maximum when \(\omega=1 / \sqrt{L C}\) ; that is, when the source frequency equals the resonance frequency of the circuit. (d) Graph \(P\) as a function of \(\omega\) for \(V=100 \mathrm{V}\) .

31\. 27. (a) Show that for an \(L-R-C\) series circuit the power factor is equal to \(R / Z\) . (Hint: Use the phasor diagram; see Fig. \(31.13 b . )\) (b) Show that for any ac circuit, not just one containing pure resistance only, the average power delivered by the voltage source is given by \(P_{\mathrm{av}}=I_{\mathrm{rms}}^{2} R\)

31.10. A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

31.28. An L-R-C series circuit is connected to a \(120-\mathrm{Hz}\) ac source that has \(V_{\mathrm{rms}}=80.0 \mathrm{V}\) . The circuit has a resistance of 75.0\(\Omega\) and an impedance at this frequency nf 105\(\Omega\) . What average power is delivered to the circuit by the source?

31.64. The Resonance width. Consider an \(L-R-C\) series circuit with a \(1.80-\mathrm{H}\) inductor, a \(0.900-\mu F\) capacitor, and a \(300-\Omega\) resistor. The source has terminal rms voltage \(V_{\mathrm{rms}}=60.0 \mathrm{V}\) and variable angular frequency \(\omega .\) (a) What is the resonance angular frequency \(\omega_{0}\) of the circuit? (b) What is the rms current through the circuit at resonance, \(I_{\max 0} ?(\mathrm{c})\) For what two values of the angular frequency, \(\omega_{1}\) and \(\omega_{2},\) is the rms current half the resonance value? (d) The quantity \(\left|\omega_{1}-\omega_{2}\right|\) defines the resonance width. Calculate \(I_{\text { rmid }}\) and the resonance width for \(R=300 \Omega, 30.0 \Omega,\) and \(3.00 \Omega .\) Describe how your results compare to the discussion in Section \(31.5 .\)
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