Calculating Current in Primary Coil: Where Does P = V^2/R Formula Come In?

In summary: Rboth forms are equivalent...if one is calculating dissipation through load resistance R its good but how you got 5 amp as current?
  • #1
Muhammad Danish
70
7

Homework Statement


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Homework Equations


How will we calculate the current in primary coil?

The Attempt at a Solution


The current in the secondary coil will be 2 Amperes by I = V/R. So in the primary coil by using ratio proportion method, the current in the primary coil will be 5 Amperes. But this leads to wrong answer.
 

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  • #2
Show work for how you come up with 5A for the primary coil current as a starting point...
 
  • #3
Muhammad Danish said:
The current in the secondary coil will be 2 Amperes by I = V/R. So in the primary coil by using ratio proportion method, the current in the primary coil will be 5 Amperes. But this leads to wrong answer.

is there a relation between Vp Vs and Ip , Is ? ultimately they are related to number of turns also.
 
  • #4
SunThief said:
Show work for how you come up with 5A for the primary coil current as a starting point...
600V ---- x
240V ---- 2
Since they are directly proportional, we can get x = 5 ampere but I don't think this is a correct answer..
 
  • #5
drvrm said:
is there a relation between Vp Vs and Ip , Is ? ultimately they are related to number of turns also.
But number of turns in primary and secondary coil are not mentioned..
 
  • #6
By law of conservation of energy - power is conserved in the input and output

Vp . Ip = Vs. Is

.
 
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  • #7
drvrm said:
By law of conservation of energy - power is conserved in the input and output

Vp . Ip = Vs. Is

.
Oh Thanks, by finding Ip I can find the power dissipated by the formula P = IV right?
 
  • #8
Muhammad Danish said:
Oh Thanks, by finding Ip I can find the power dissipated by the formula P = IV right?
Just to increase knowledge, where do we use the formula P = V^2/R ?
 
  • #9
Muhammad Danish said:
Just to increase knowledge, where do we use the formula P = V^2
R
its v^2/R
both forms are equivalent...if one is calculating dissipation through load resistance R its good
but how you got 5 amp as current?.
 

1. What is a Step-Down Transformer?

A step-down transformer is a type of electrical transformer that has a higher number of windings on the primary coil than on the secondary coil. This allows it to decrease the voltage from the primary source to a lower voltage on the secondary side.

2. How does a Step-Down Transformer work?

A step-down transformer works on the principle of electromagnetic induction. When an alternating current flows through the primary coil, it creates a changing magnetic field. This magnetic field then induces a voltage in the secondary coil, which is lower than the primary voltage due to the difference in the number of windings.

3. What is the purpose of a Step-Down Transformer?

The main purpose of a step-down transformer is to reduce the voltage of an alternating current. This is useful in many different applications, such as in household appliances where a lower voltage is required for safety reasons, or in power distribution systems where high voltage transmission is more efficient.

4. How is the voltage ratio determined in a Step-Down Transformer?

The voltage ratio in a step-down transformer is determined by the ratio of the number of windings on the primary coil to the number of windings on the secondary coil. For example, if the primary coil has 100 windings and the secondary coil has 50 windings, the voltage will be halved on the secondary side.

5. What are some common uses of Step-Down Transformers?

Step-down transformers are commonly used in power adapters for electronic devices, such as laptops and cell phones, to convert the high voltage from a wall outlet to a lower voltage that can be safely used by the device. They are also used in power distribution systems to step down the high voltage from power plants to a lower voltage for household use. Additionally, they are used in various industrial and commercial applications to power equipment that requires lower voltages.

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