Question

An ideal transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary? And why?

Answer 1

The current in the primary coil of an ideal transformer with a 500-turn primary coil and a 2000-turn secondary coil, when the current in the secondary is 3.0 A, is 12 A.

Step-by-step explanation:

An ideal transformer is a device that can change the voltage and current levels of an alternating current (AC) signal. In this case, we have a transformer with a primary coil of 500 turns and a secondary coil of 2000 turns. The current in the secondary coil is 3.0 A. To find the current in the primary coil, we can use the formula:

Np/Ns = Ip/Is

where Np is the number of turns in the primary coil, Ns is the number of turns in the secondary coil, Ip is the current in the primary coil, and Is is the current in the secondary coil.

Plugging in the values, we get:

Ip/3.0 = 500/2000

Simplifying the equation gives:

Ip = 12 A

Therefore, the current in the primary coil is 12 A.

Answer 2

12 A

Step-by-step explanation:

`N_p` = Number of turns in the primary coil = 500

`N_s` = Number of turns in the secondary coil = 2000

`I_p` = Current in the primary coil

`I_s` = Current in the secondary coil = 3 A

Transformers follow the equation

`(N_p)/(N_s)=(I_s)/(I_p)\\\Rightarrow I_p=(I_s* N_s)/(N_p)\\\Rightarrow I_p=(3* 2000)/(500)\\\Rightarrow I_p=12\ A`

The current in the primary coil is 12 A

An ideal transformer follows the above equation