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Chapter 33: Problem 46

A step-up transformer is designed to have an output voltage of \(2200 \mathrm{V}\) (rms) when the primary is connected across a \(110-\mathrm{V}\) (rms) source. (a) If the primary winding has 80 turns, how many turns are required on the secondary? (b) If a load resistor across the secondary draws a current of \(1.50 \mathrm{A},\) what is the current in the primary, assuming ideal conditions? (c) What If? If the transformer actually has an efficiency of \(95.0 \%,\) what is the current in the primary when the secondary current is \(1.20 \mathrm{A} ?\)

Short Answer

Expert verified
The number of turns required on the secondary are \(1599.9\) turns. The current in the primary, under ideal conditions, is \(30A\). Considering an efficiency of \(95% \), the current in the primary when the secondary current is \(1.20A\) is \(26.6A\).

Step by step solution

01

Find secondary turns

For a transformer, the ratio of secondary to primary voltages is equal to the ratio of secondary to primary turns. So, we calculate the secondary turns using the formula \(N_s = (V_s/V_p) * N_p\), where \(N_p\) is the primary turns, \(N_s\) is the secondary turns, \(V_p\) is the primary voltage and \(V_s\) is the secondary voltage. We insert the given values, \(V_s = 2200V\), \(V_p = 110V\) and \(N_p = 80\) into the equation to calculate \(N_s\).
02

Find primary current

For an ideal transformer, the power input to the primary coil is equal to the power output from the secondary coil. So, the product of the primary voltage and current is equal to the product of the secondary voltage and current. From this, we can deduce that the primary current \(I_p\) is equal to \(I_s * (V_s/V_p)\) where \(I_s\) is the secondary current and \(V_s\) and \(V_p\) are again the voltages at the secondary and primary coils. We plug in the given values, \(I_s = 1.50A\), \(V_s = 2200V\), and \(V_p = 110V\) to calculate the primary current.
03

Account for efficiency

Considering the efficiency of the transformer, we calculate the actual current entering the transformer at the primary coil as \(I_p = I_s * (V_s/V_p)/ (\eta/100)\), with the efficiency \(\eta = 95.0% \) being given. We take \( I_s = 1.20A \), and calculate the primary current, accounting for the efficiency of the transformer.
04

Finalize your answers

After calculations, finalize your answers and cross verify for any calculation or conceptual mistakes.

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Key Concepts

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

Understanding Step-up Transformers
Step-up transformers are fascinating devices that play a crucial role in increasing electrical voltage levels. They operate based on the principle of electromagnetic induction, allowing energy to be transferred between two circuits through electromagnetic fields.
  • A step-up transformer has more turns on the secondary coil, compared to the primary.
  • This configuration effectively increases the voltage from the primary side to the secondary side.
  • The voltage ratio is directly related to the number of turns on each coil, stated as \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \).
In our example, a transformer designed for 110V input and 2200V output requires more turns on the secondary coil. Using the given formula, you can determine the exact number of turns needed on the secondary for proper voltage transformation.

Calculating the number of secondary turns involves simple multiplication: you multiply the turn ratio by the primary coil's turns to find the secondary coil's turns, ensuring the transformer performs efficiently in its task of stepping up voltage.
Current Calculations in Transformers
Calculating current in transformers involves understanding the relationship between voltage and current in an electromagnetic circuit. For an ideal transformer:
  • The power input to the primary coil equals the power output from the secondary coil. Hence, \( V_p \times I_p = V_s \times I_s \).
  • This relation allows us to calculate the primary current when the secondary current is known, using \( I_p = I_s \times (\frac{V_s}{V_p}) \).
In an example where the transformer has a secondary current \( I_s \) of 1.50A, we substitute the provided voltages into the equation to calculate the primary current.

Transformers work on the principle of inverse proportionality. As the voltage increases, the current decreases, and vice versa. It's crucial to understand these calculations help maintain the energy conservation within transformer circuits, assuring no power is unaccounted for in ideal conditions.
Exploring Transformer Efficiency
Transformer efficiency is critical to ensure that energy losses are minimized during the voltage transformation process. While ideal transformers assume 100% efficiency, real-world applications often see slight losses due to factors like heat dissipation and resistance.
  • Efficiency is defined as the ratio of output power to input power, often expressed as a percentage.
  • In this context, we consider a transformer's efficiency to be 95.0%, meaning it effectively delivers 95% of its input power to its load.
  • To calculate the primary current with given efficiency, use the formula \( I_p = \frac{I_s \times (\frac{V_s}{V_p})}{\frac{\eta}{100}} \).
For example, given a secondary current of 1.20A and a 95% efficiency, calculating the primary current includes dividing by the efficiency percentage.

Efficiency calculations like these are essential in real-world applications to ensure that the transformers are operating optimally and delivering as much power as possible with minimal loss.

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Most popular questions from this chapter

A transmission line that has a resistance per unit length of \(4.50 \times 10^{-4} \Omega / \mathrm{m}\) is to be used to transmit \(5.00 \mathrm{MW}\) over 400 miles \(\left(6.44 \times 10^{5} \mathrm{m}\right) .\) The output voltage of the generator is \(4.50 \mathrm{kV} .\) (a) What is the line loss if a transformer is used to step up the voltage to \(500 \mathrm{kV} ?\) (b) What fraction of the input power is lost to the line under these circumstances? (c) What If? What difficulties would be encountered in attempting to transmit the \(5.00 \mathrm{MW}\) at the generator voltage of \(4.50 \mathrm{kV} ?\)

To determine the inductance of a coil used in a research project, a student first connects the coil to a \(12.0-\mathrm{V}\) battery and measures a current of \(0.630 \mathrm{A}\). The student then connects the coil to a \(24.0-\mathrm{V}\) (rms), \(60.0-\mathrm{Hz}\) generator and measures an rms current of \(0.570 \mathrm{A}\). What is the inductance?

A series \(R L C\) circuit in which \(R=1.00 \Omega, L=\) \(1.00 \mathrm{mH},\) and \(C=1.00 \mathrm{nF}\) is connected to an \(\mathrm{AC}\) source delivering \(1.00 \mathrm{V}\) (rms). Make a precise graph of the power delivered to the circuit as a function of the frequency and verify that the full width of the resonance peak at half-maximum is \(R / 2 \pi L.\)

Suppose you manage a factory that uses many electric motors. The motors create a large inductive load to the electric power line, as well as a resistive load. The electric company builds an extra-heavy distribution line to supply you with a component of current that is \(90^{\circ}\) out of phase with the voltage, as well as with current in phase with the voltage. The electric company charges you an extra fee for "reactive volt-amps," in addition to the amount you pay for the energy you use. You can avoid the extra fee by installing a capacitor between the power line and your factory. The following problem models this solution. In an \(R L\) circuit, a \(120-\mathrm{V}\) (rms), \(60.0-\mathrm{Hz}\) source is in series with a 25.0 -mH inductor and a \(20.0-\Omega\) resistor. What are (a) the rms current and (b) the power factor? (c) What capacitor must be added in series to make the power factor \(1 ?\) (d) To what value can the supply voltage be reduced, if the power supplied is to be the same as before the capacitor was installed?

One particular plug-in power supply for a radio looks similar to the one shown in Figure 33.23 and is marked with the following information: Input \(120 \mathrm{V}\) AC \(8 \mathrm{W}\) Output \(9 \mathrm{V}\) DC 300 mA. Assume that these values are accurate to two digits. (a) Find the energy efficiency of the device when the radio is operating. (b) At what rate does the device produce wasted energy when the radio is operating? (c) Suppose that the input power to the transformer is \(8.0 \mathrm{W}\) when the radio is switched off and that energy costs \(\$ 0.135 / \mathrm{kWh}\) from the electric company. Find the cost of having six such transformers around the house, plugged in for thirty- one days.
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