Alternating Current Circuits and Electromagnetic Waves

In summary, the conversation discusses the use of a step-down transformer for recharging batteries in portable devices. The transformer has a turns ratio of 13:1 and is used with 120V household service. The question is posed about the voltage and current supplied to a tape player from the transformer, as well as the power delivered. The conversation then elaborates on the properties of an ideal transformer and how they relate to the given scenario.
  • #1
ScoutFCM
17
0
Just wanted to know if someone could help me out or check my answers.

The Transformer
38) A step-down transformer is used for recharging the batteries of portable devices such as tape players. The turns ratio inside the transformer is 13:1 and it is used with 120V (rms) household service. If a particular ideal transformer draws 0.350A from the house outlet, what are (a) the voltage and (b) the current supplied to a tape player from the transformer? (c) How much power is delivered?

a) ∆V= -N(∆Ф)/∆t = ?
I don't know how to get current/number of turns.

b) Ф= V∆t/-N ?

c) (.350A)(120V) = 42W
 
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  • #2
properties of an ideal transformer

You need to understand two things about an ideal transformer:
(1) The ratio of the voltages (primary to secondary) equals the ratio of the number of turns. (This comes from Faraday's law.)
(2) The power in the primary equals the power in the secondary (from energy conservation):
[tex]V_pI_p = V_sI_s[/tex]
 
  • #3


(a) To find the output voltage, we can use the turns ratio of the transformer. Since the transformer is a step-down transformer with a turns ratio of 13:1, the output voltage will be 1/13th of the input voltage. Therefore, the output voltage to the tape player will be 120V/13 = 9.23V (rms).

(b) The current supplied to the tape player can be found using Ohm's Law, where I = V/R. We know the output voltage (9.23V) and the resistance of the tape player (unknown), so we cannot calculate the current supplied to the tape player without more information.

(c) Power can be calculated using the formula P = VI. Since we know the output voltage (9.23V) and the current drawn from the house outlet (0.350A), we can calculate the power delivered to the tape player as (9.23V)(0.350A) = 3.23W.
 

FAQ: Alternating Current Circuits and Electromagnetic Waves

What is alternating current (AC) and how does it differ from direct current (DC)?

Alternating current is a type of electrical current where the direction of flow of electricity constantly changes. This is in contrast to direct current, where the flow of electricity is unidirectional. AC is typically used in power distribution systems, while DC is commonly used in electronic devices.

How do alternating current circuits work?

In an AC circuit, the electricity flows back and forth between the two terminals of a power source, creating a cycle of alternating current. This is achieved by using alternating current generators, which use rotating magnets to induce an alternating current in a wire coil.

What are the advantages of using alternating current?

One of the main advantages of AC is that it can be easily transformed into different voltages using transformers, making it efficient for long-distance power transmission. Additionally, AC can power multiple devices simultaneously, whereas DC can only power one device at a time.

What are electromagnetic waves and how are they related to AC circuits?

Electromagnetic waves are a form of energy that travels through space in the form of oscillating electric and magnetic fields. AC circuits produce electromagnetic waves, which can then be used for communication and other applications.

What are some common applications of alternating current circuits and electromagnetic waves?

AC circuits and electromagnetic waves have a wide range of applications, including power generation and distribution, electric motors, radio and television broadcasting, wireless communication, and many more. They are also used in various medical devices, such as MRI machines, to produce electromagnetic fields for imaging purposes.

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