Primary coil of an ideal transformer. I need an equation.

In summary, the primary coil of an ideal transformer is connected to a 120-V source and draws 1.0 A, while the secondary coil has 600 turns and supplies an output current of 5.0 A. The voltage across the secondary coil is 24 volts and there are 3000 turns in the primary coil. To find the maximum input current for the transformer, the load resistance (R) is calculated using the voltage and current values from the secondary coil, and then used to find the secondary voltage (Vs). Using the given maximum power (P) of 210 watts, the secondary current (Is) is then calculated and used to find the input current (Ip) using the equation for the transformer turns ratio.
  • #1
Sarak
3
0

Homework Statement


The primary coil of an ideal transformer is connected to a 120-V source and draws 1.0 A. The secondary coil has 600 turns and supplies an output current of 5.0 A to run an electrical device.
A) What is the voltage across the secondary coil?
B) How many turns are in the primary coil?
C) If the maximum power allowed by the device (before it is destroyed) is 210 W, what is the maximum input current to this transformer?


Homework Equations


V2=V1I1/I2
N1=N2I2/I1
and ? I=P/V ?


The Attempt at a Solution


I found part A) to be V2=24, and part B) to be N1=3000. The only equation I can think of for part C) is I=P/V. P is given to be 210 W. I tried plugging V1 and V2 in that equation and haven't gotten the right answer. Can someone given me an idea how to go about this? I can't find anything in the text.
 
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  • #2
Maybe you won't see this reply or no more need help but let me reply.

Let's find R(load resistance) from first,

R=V/I = 24/5 ohm

and

Vs= Is*R = Is * 24/5 Volt

also

Ps=Pp= 210 watt

lets use these,

Ps= Vs*Is = Is * 24/5 * Is = 210 watt

Is^2 = 210 * 5/24 A

Is = 6.61 A

now let's find input current. From

Ip/Is = Ns/Np

and

Ip=Ns/Np * Is = 600/3000 * 6.61 = 1,322 A

I suppose approx. correct.
 

Related to Primary coil of an ideal transformer. I need an equation.

1. What is the purpose of the primary coil in an ideal transformer?

The primary coil of an ideal transformer is responsible for receiving and transmitting electrical energy. It is connected to the input voltage source and is designed to produce a changing magnetic field.

2. What is the difference between the primary coil and secondary coil in an ideal transformer?

The primary coil has a larger number of turns and is connected to the input voltage source, while the secondary coil has a smaller number of turns and is connected to the output load. The primary coil also produces a changing magnetic field, while the secondary coil receives this field and converts it into electrical energy for the output.

3. How is the primary coil connected in an ideal transformer?

The primary coil is connected to the input voltage source through a power supply and is designed to have a higher voltage than the secondary coil. It is typically made of copper wire and is wound around an iron core to increase the strength of the magnetic field.

4. What is the equation for the primary coil in an ideal transformer?

The equation for the primary coil in an ideal transformer is V1 = N1/N2 * V2, where V1 is the input voltage, N1 is the number of turns in the primary coil, V2 is the output voltage, and N2 is the number of turns in the secondary coil.

5. How does the primary coil affect the efficiency of an ideal transformer?

The primary coil plays a crucial role in the efficiency of an ideal transformer. A well-designed primary coil with a high number of turns and a proper connection to the input voltage source can increase the efficiency of the transformer by minimizing energy losses and maximizing the strength of the magnetic field.

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