=15.3/18 x 100

In the solution to the worked example 2, the efficiency of the electric motor is stated to be 60%. (a) What is the ratio of turns in the new secondary compared with the old secondary? (a) What is the voltage output of a transformer used for rechargeable flashlight batteries, if its primary has 500 turns, its secondary 4 turns, and the input voltage is 120 V? The magnetic field created by the primary is mostly confined to and increased by the core, which transmits it to the secondary coil. Thanks for the heads-up. 4. 9. A typical construction of a simple transformer has two coils wound on a ferromagnetic core that is laminated to minimize eddy currents. A step-up transformer is one that increases voltage, whereas a step-down transformer decreases voltage. Not only does the iron core trap the magnetic field created by the primary coil, its magnetization increases the field strength. Power is the rate that work is done, and when you exercise you do As given in the question, the efficiency of the electric motor is 40%. (a) What is the ratio of turns in the primary and secondary coils of her transformer? Transformers have many applications in electrical safety systems, which are discussed in Electrical Safety: Systems and Devices. about 10 Newtons.

Convert Power into Total input Energy using formulae. By the end of this section, you will be able to: Transformers do what their name implies—they transform voltages from one value to another (The term voltage is used rather than emf, because transformers have internal resistance). A pendulum Bob swings from one end to the other.

2. The type of transformer considered in this text—see Figure 3—is based on Faraday’s law of induction and is very similar in construction to the apparatus Faraday used to demonstrate magnetic fields could cause currents.

The lamp wasted 2.7J of energy in 3.0s. What are the numbers of turns in the parts of the secondary used to produce the output voltages?

Recall that power is given by $P = \frac{W}{t}$. When the useful energy (7.2 x 10^5) is divided by the total energy (1.8 x 10^6), it amounts to 0.4.

(a) How many loops should there be in the secondary coil? Keep it up! If you spot any errors or want to suggest improvements, please contact us. A softballer throws a ball into the air and catches it on the way down.state the energy changes that take place ? A large number of loops in the secondary (compared with the primary) is required to produce such a large voltage. A large power plant generates electricity at 12.0 kV.

1. Since 60% of the energy from the electric motor is lost as heat and sound, only 40% is useful energy output.

This gives.

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Discuss the possibility of power losses in the devices and the effect on the output current and power. Subtract your answer from Wasted Energy , you will get you Useful Energy.Then use Efficiency formulae to solve your answer by dividing useful energy by total energy multiply by 100.

(b) What is its input current when its output is 1.30 A? A portable x-ray unit has a step-up transformer, the 120 V input of which is transformed to the 100 kV output needed by the x-ray tube.

where Vp and Vs are the voltages across primary and secondary coils having Np and Ns turns.

Suppose you lift the book 1 meter, 10 times in 20 In normal use, the input voltage is placed on the primary, and the secondary produces the transformed output voltage. (c) How could a New Zealander traveling in the United States use this same transformer to power her 240 V appliances from 120 V? In certain spectacular demonstrations, very large voltages are used to produce long arcs, but they are relatively safe because the transformer output does not supply a large current. Many common exercises involve doing work. For instance, each time you do a pushup, you lift the weight of

Assume 100% efficiency. you will measure distances, forces, and times in order to calculate

Some transformers even provide a variable output by allowing connection to be made at different points on the secondary coil.

But high voltages pose greater hazards, so that transformers are employed to produce lower voltage at the user’s location. Power tells us how fast work is being done or how fast energy is being converted from one form to another.

A power supply is also known as a power supply unit, power brick or power adapter.

We also see that a small input current produces a larger output current in a step-down transformer. [latex]\begin{array}{lll}{N}_{\text{s}}& =& {N}_{\text{p}}\frac{{V}_{\text{s}}}{{V}_{\text{p}}}\\ & =& \left(50\right)\frac{100,000\text{ V}}{120\text{ V}}=4.17\times 10^{4}\end{array}\\[/latex]. I have a question J, Power output = (Total work)/(time) = 100 J / 20 s = 5 Watts. This would be true for neon sign transformers and those supplying high voltage inside TVs and CRTs. number of repetitions that you performed. Faraday’s law of induction for the secondary coil gives its induced output voltage Vs to be. is the relationship between the output and input currents of a transformer. Figure 3.

Click to download the simulation. A common physics lab involves quickly climbing a flight of stairs and using mass, height and time information to determine a student's personal power. This is known as the transformer equation, and it simply states that the ratio of the secondary to primary voltages in a transformer equals the ratio of the number of loops in their coils.

SI Unit for power is watt (W), scalar quantity.

total time taken.

, where F = force, v = velocity (Simple derivation below). (a) The plug-in transformer for a laptop computer puts out 7.50 V and can supply a maximum current of 2.00 A.

[latex]{V}_{\text{s}}=-{N}_{\text{s}}\frac{\Delta\Phi}{\Delta t}\\[/latex], where Ns is the number of loops in the secondary coil and ΔΦ/Δt is the rate of change of magnetic flux. you exerted the force.

What is the efficiency of the lamp? Made with | 2010 - 2020 | Mini Physics |, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on WhatsApp (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Pocket (Opens in new window), Click to share on Skype (Opens in new window), Case Study 1: Energy Conversion for An Oscillating Ideal Pendulum, Case Study 2: Energy Conversion for A Bouncing Ball, O Level: Magnetic Field And Magnetic Field Lines, One watt (W) is defined as the rate of work done or, $\text{One watt} = \frac{\text{one joule}}{\text{one second}}$ $\rightarrow$ $1 \, \text{W} = 1 \, \text{J s}^{-1}$.

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