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The rechargeable batteries for a laptop computer need a much smaller voltage than what a wall socket provides. Therefore, a transformer is plugged into the wall socket and produces the necessary voltage for charging the batteries. The batteries are rated at \(9.0 \mathrm{V},\) and a current of \(225 \mathrm{mA}\) is used to charge them. The wall socket provides a voltage of \(120 \mathrm{V}\). (a) Determine the turns ratio of the transformer. (b) What is the current coming from the wall socket? (c) Find the average power delivered by the wall socket and the average power sent to the batteries.

Short Answer

Expert verified
(a) 13.33; (b) 16.9 mA; (c) 2.028 W, 2.025 W.

Step by step solution

01

Understand the Problem

The problem deals with determining various parameters of a transformer used for charging laptop batteries with given specifications. We need to calculate the transformer's turns ratio, the current coming from the wall socket, and the average power delivered to and received from the wall socket.
02

Calculate the Turns Ratio

The turns ratio of the transformer is given by the ratio of the input voltage to the output voltage. We use the formula: \[ \text{Turns Ratio (n)} = \frac{V_{input}}{V_{output}} \]Substitute the given values: \[ n = \frac{120\, \text{V}}{9.0\, \text{V}} \]Therefore, the turns ratio is approximately 13.33.
03

Calculate the Current from the Wall Socket

The power in the primary and secondary coils of a transformer (ignoring losses) is equal: \( P_{primary} = P_{secondary} \). We have:\[ V_{input} \times I_{primary} = V_{output} \times I_{output} \]Solving for \( I_{primary} \), we get:\[ I_{primary} = \frac{V_{output} \times I_{output}}{V_{input}} \]Substitute the given values:\[ I_{primary} = \frac{9.0\, \text{V} \times 225\, \text{mA}}{120\, \text{V}} = \frac{9.0\, \text{V} \times 0.225\, \text{A}}{120\, \text{V}} \]The current from the wall socket is approximately 0.0169 A (16.9 mA).
04

Find the Average Power Delivered by the Wall Socket

The power delivered by the wall socket is given by:\[ P_{primary} = V_{input} \times I_{primary} \]Substitute the known values:\[ P_{primary} = 120\, \text{V} \times 0.0169\, \text{A} \approx 2.028\, \text{W} \]
05

Find the Power Sent to the Batteries

The power sent to the batteries is given by:\[ P_{secondary} = V_{output} \times I_{output} \]Substitute the known values:\[ P_{secondary} = 9.0\, \text{V} \times 0.225\, \text{A} = 2.025\, \text{W} \]
06

Conclusion: Summarize the Findings

The turns ratio of the transformer is approximately 13.33. The current coming from the wall socket is approximately 16.9 mA. The average power delivered by the wall socket is approximately 2.028 W, and the average power sent to the batteries is approximately 2.025 W.

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Key Concepts

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

Turns Ratio
In transformers, the turns ratio is a crucial concept and determines how the voltages are transformed from one side to the other. The turns ratio is the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. This ratio dictates how the input voltage is stepped-up or stepped-down to meet the required output voltage level. For example, if you have a transformer connected to a wall socket with 120 V, and the equipment (like a laptop battery) requires only 9 V, the turns ratio helps adjust the voltage to a suitable level. You calculate it using the formula:
  • Turns Ratio (n) = \(\frac{V_{\text{input}}}{V_{\text{output}}}\)
When you insert the necessary values:
  • \(n = \frac{120\, \text{V}}{9.0\, \text{V}}\)
You’ll find that the turns ratio is approximately 13.33. This means the primary coil has 13.33 times more turns than the secondary coil to achieve the required voltage reduction.
Current Calculation
The current from the wall socket, or the input current (\(I_{\text{primary}}\)), can be determined by understanding the conservation of power in transformers. Despite some power loss in practical applications, transformers ideally ensure that power in the primary circuit equals the power in the secondary circuit, granted the assumption of no losses: \(P_{\text{primary}} = P_{\text{secondary}}\).To calculate the input current, remember the relationship:
  • \(V_{\text{input}} \times I_{\text{primary}} = V_{\text{output}} \times I_{\text{output}}\)
Solving for the primary current:
  • \(I_{\text{primary}} = \frac{V_{\text{output}} \times I_{\text{output}}}{V_{\text{input}}}\)
Substitute the known values of 9.0 V and 225 mA:
  • \(I_{\text{primary}} = \frac{9.0\, \text{V} \times 0.225\, \text{A}}{120\, \text{V}}\)
This calculation shows that the current coming from the wall socket is approximately 16.9 mA. Knowing this current is important for managing energy efficiency and ensuring safety when designing or using electronic devices.
Power Delivered
The power delivered through the transformer from the wall socket is a measurement of how effectively energy is transmitted from the primary side to the secondary side. It involves calculating the total energy per unit time delivered to the primary and secondary circuits.The power delivered by the wall socket, represented as primary power \(P_{\text{primary}}\), uses the equation:
  • \(P_{\text{primary}} = V_{\text{input}} \times I_{\text{primary}}\)
By inserting the values:
  • \(P_{\text{primary}} = 120\, \text{V} \times 0.0169\, \text{A} \approx 2.028\, \text{W}\)
Similarly, the power used by the batteries, or the secondary power \(P_{\text{secondary}}\), is calculated as:
  • \(P_{\text{secondary}} = V_{\text{output}} \times I_{\text{output}}\)
  • \(P_{\text{secondary}} = 9.0\, \text{V} \times 0.225\, \text{A} = 2.025\, \text{W}\)
Understanding both ends of power usage gives insights on the efficiency and effectiveness of the transformer in serving its purpose while minimizing energy waste. This knowledge helps in optimizing performance and ensuring devices function within safe and reliable electrical parameters.

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Most popular questions from this chapter

Consult Multiple-Concept Example 11 for background material relating to this problem. A small rubber wheel on the shaft of a bicycle generator presses against the bike tire and turns the coil of the generator at an angular speed that is 38 times as great as the angular speed of the tire itself. Each tire has a radius of \(0.300 \mathrm{m}\). The coil consists of 125 turns, has an area of \(3.86 \times 10^{-3} \mathrm{m}^{2},\) and rotates in a \(0.0900-\mathrm{T}\) magnetic field. The bicycle starts from rest and has an acceleration of \(+0.550 \mathrm{m} / \mathrm{s}^{2} .\) What is the peak emf produced by the generator at the end of 5.10 s?

A Generator Bike. You and your team are designing a generator using a stationary bike to rotate a coil in a uniform magnetic field. The gearing is set up so that the coil rotates 60 times for one complete rotation of the bike pedals. Therefore, one revolution of the pedals per second results in a \(60-\mathrm{Hz}\) alternating current in the coil. The circular coil has 350 turns and a diameter of \(15.0 \mathrm{cm},\) and its axis of rotation is along its diameter. (a) If a uniform magnetic field is oriented perpendicular to the coil's axis of rotation and has a magnitude of \(B=0.225 \mathrm{T}\), what is the peak emf produced by the generator bike? (b) What is the rms emf? (c) To what magnitude should you reduce the field if you want the rms emf to be 110 VAC? (d) Instead of reducing the field, you could use a step-down transformer to reduce the rms emf to 110 VAC. What should be the ratio of primary to secondary turns of the transformer coils?

A generator has a square coil consisting of 248 turns. The coil rotates at \(79.1 \mathrm{rad} / \mathrm{s}\) in a \(0.170-\mathrm{T}\) magnetic field. The peak output of the generator is \(75.0 \mathrm{V}\). What is the length of one side of the coil?

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3} \mathrm{Wb}\). What is the flux that passes through the circular loop?

A constant current of \(I=15 \mathrm{A}\) exists in a solenoid whose inductance is \(L=3.1 \mathrm{H}\). The current is then reduced to zero in a certain amount of time. (a) If the current goes from 15 to \(0 \mathrm{A}\) in a time of \(75 \mathrm{ms},\) what is the emf induced in the solenoid? (b) How much electrical energy is stored in the solenoid? (c) At what rate must the electrical energy be removed from the solenoid when the current is reduced to \(0 \mathrm{A}\) in a time of \(75 \mathrm{ms}\) ? Note that the rate at which energy is removed is the power.

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