Transformer power losses at no-loading conditions (core or iron losses)

<div dir="ltr" style="text-align: left;" trbidi="on"><br /> Hysterics core losses<br /> <br /> 1.Magnetic energy stored during build-up of magnetic field greater than magnetic energy retrieved during decay of magnetic field.<br /> 2.The difference is known as hysterics energy losses <br /> <br /> <img alt="http://img103.herosh.com/2010/12/05/255094724.png" height="508" src="http://img103.herosh.com/2010/12/05/255094724.png" style="cursor: -moz-zoom-in;" width="718" /><br /> <br /> 2.Eddy currents core losses:<br />   <br /> <br /> Induced eddy currents  can be minimized <br /> <br /> by using laminated sheet steel magnetic core type<br /> <br /> <img alt="http://img101.herosh.com/2010/12/05/548789843.png" src="http://img101.herosh.com/2010/12/05/548789843.png" /><br /> <br /> </div>

• 
• 

1. 1
2. 2

• Previous
• Next

Transformer power losses at no-loading conditions (core or iron losses)

Hysterics core losses 1.Magnetic energy stored during build-up of magnetic field greater than magnetic energy retrieved during decay of magnetic field. 2.The difference is known as hysterics energy l…

more information

Transformer on no-loading condition

<div dir="ltr" style="text-align: left;" trbidi="on"><b><br /> 1.Transformer on no-loading condition<br /> <br /> ØNo-load voltage ratio<br /> <br /> ØCore losses: hysterisis & eddy current losses <br /> <br /> ØSimulation circuit <br /> <br /> 2.Simulation (or equivalent) circuit at loading condition </b></div>

Transformer on no-loading condition

1.Transformer on no-loading condition ØNo-load voltage ratio ØCore losses: hysterisis & eddy current losses ØSimulation circuit 2.Simulation (or equivalent) circuit at loading condition …

more information

Solved Problem

<div dir="ltr" style="text-align: left;" trbidi="on"><span style="font-size: large;">Consider the circuit showing in the figure. Assume that the core has a relative permeability of 5000, with a rectangular cross section of 2cm by 2cm and that the wire has 300 turns. Determine the current required to establish a flux density in the air gap, Bgap of 0.5 Tesla. </span><div class="O"><div style="text-align: justify;"></div></div></div>

Solved Problem

Consider the circuit showing in the figure. Assume that the core has a relative permeability of 5000, with a rectangular cross section of 2cm by 2cm and that the wire has 300 turns. Determine the curr…

more information

Magnetic Circuits part 2

<div dir="ltr" style="text-align: left;" trbidi="on"><div style="color: blue;"><a href="http://hayatei.blogspot.com/2010/12/magnetic-circuits_04.html"><b><span style="font-size: large;">Magnetic Circuits part 1</span></b></a></div><div style="color: red;"><b><span style="font-size: large;"><br /> </span></b></div><div style="color: red;"><b><span style="font-size: large;">3.Ohm’s Law of Electromagnetism</span></b></div><div style="color: red;"><b><span style="font-size: large;"><br /> </span></b></div><div style="color: red;"><b><span style="font-size: large;">From the given magnetic circuit</span></b></div><div style="color: red;"><b><span style="font-size: large;"><br /> </span></b></div><div style="color: red;"><b><span style="font-size: large;">F = B × A = m × B × A</span></b></div><div style="color: red;"><b><span style="font-size: large;"><br /> </span></b></div><div style="color: red;"><b><span style="font-size: large;">From Ampere’s circuital law</span></b></div><div style="color: red;"><b><span style="font-size: large;"><br /> </span></b></div><div style="color: red;"><b><span style="font-size: large;"><br /> </span></b></div><div style="color: red;"><b><span style="font-size: large;"> h=ni/i</span></b></div><div style="color: red;"><b><span style="font-size: large;"><br /> </span></b></div><div style="color: red;"><b><span style="font-size: large;">FMMF  (NI) adds or subtracts around a loop like voltage sources</span></b></div><div style="color: red;"><b><span style="font-size: large;"> in an electric circuit.F Reluctance like resistance can be in series or in parallel.</span></b></div><div style="color: red;"><b><span style="font-size: large;">F F divides in parallel paths in inverse proportion of the reluctance like current with </span></b></div><div style="color: red;"><b><span style="font-size: large;">parallel resistors</span></b></div></div>

Magnetic Circuits part 2

Magnetic Circuits part 1 3.Ohm’s Law of Electromagnetism From the given magnetic circuit F = B × A = m × B × A From Ampere’s circuital law  h=ni/i FMMF  (NI) adds or subtracts around a loop like volta…

more information