Battery with voltage V=12V is connected to the primary of a transform

[ThirdEyeBlind]

Homework Statement

A battery with voltage V=12V is connected to the primary of a transformer. A resistor with resistance R =20 Ohms is connected to the transformer secondary. The transformer has an iron core with Np=200 turns on the primary and Ns=600 turns on the secondary. Assume the resistance of the wires is zero.

What is Is, the current in the resistor?

Homework Equations

Vs= (Ns/Np)Vp
V=IR

The Attempt at a Solution

Vs=(600/200) * 12 V
Vs=36 V

V=IR

36V=I(20 ohms)
I =1.8A, but the answer is 0? Could someone please show me where I have made my error or how to arrive at the answer of 0 A?

 

[gneill]

Transformers require AC to operate. A DC voltage source on the primary will just run a DC current through the primary (limited by the primary's resistance). After any initial transient response when the voltage is first connected the current in the primary will be constant, so there will be no changing magnetic field to influence the secondary. So after an initial 'glitch' there will be no current in the secondary.

 

[Delphi51]

The answer dies not lie in calculations.
Do you know any of the electromagnetic theory behind a transformer?
What is it that causes a current in the secondary winding?
What is the nature of the current in the primary that causes that?

 

[ThirdEyeBlind]

Transformers require AC to operate. A DC voltage source on the primary will just run a DC current through the primary (limited by the primary's resistance). After any initial transient response when the voltage is first connected the current in the primary will be constant, so there will be no changing magnetic field to influence the secondary. So after an initial 'glitch' there will be no current in the secondary.

Thank you for the quick response and explanation. I now understand.

 

[asteriatic]

Based on the given information, the current in the resistor should be 1.8A. This can be calculated using Ohm's Law, V=IR, where V is the voltage across the resistor (36V) and R is the resistance (20 ohms). Therefore, I = V/R = (36V)/(20 ohms) = 1.8A.

It is possible that the answer of 0A is referring to the current in the primary of the transformer. In this case, we can use the equation V=IR, where V is the voltage of the battery (12V) and R is the resistance of the primary circuit (zero, assuming the wires have zero resistance). Therefore, the current in the primary would be 0A.

In summary, the current in the resistor is 1.8A, while the current in the primary circuit is 0A. It is important to carefully read and understand the problem and what is being asked for in order to arrive at the correct solution.