A little help please ....

131

02:25:47

This is a question I got off my text it asks the voltage across the "8 ohm" resistor on the top of the schematics.. I've tried to solve it many times using every possible way I can think of but the answer I get just doesn't match with the answer given .. The answer is 24 volts. I tried to check it with every circuit, PSPICE, Multisim but none shows 24 V ..... What I'm I doing wrong here ...

published 11 years ago

ETJAKEOC

11 years ago

17.4V my friend, it's the higher voltage on one side of the resistor minus the other

selman

11 years ago

Yes i got that but the text says 24 V and pspice and multisim shows it is about 10.93 V

hurz

11 years ago

Strange circuit, or better said strange textbook, what is the purpose of this circuit if any? It very much depend on the transformer impedance. And which one of the "8Ohm" you mean?

hurz

11 years ago

2 ways i see to get exactly 24V. 1) Make the transformer "invisible" by increase the frequency to a high one like 10MHz. 2) remove the transformer and you get the theoretical 24V. I both cases just by ohms law

selman

11 years ago

Thanks hurz . the text book says the transformer is ideal which gives it infinite inductance and inturn infinite impedance which causes all the source current to go through the 8 ohm resistor instead of the transformer ......

hurz

11 years ago

Hmmm, perfect for me would mean infinite input impedance and zero output- Which could be simulated with an opamp or VCVS. If ouput is suppose to be also infinite in impedance than its like i said before it would be invisible which you can simlute by increase it by frequency or inductivity or just remove it.

jeysoncruz

10 years ago

Make the Coupling Coefficient (transformer) equal to 1k (999m -> 1k), maximum accepted value. Elevando o coeficiente de acoplamento de 999m para 1k, temos algo mais próximo de um transformador ideal.

jason9

10 years ago

The transformer is backwards.

kiani

7 years ago

i think for start reduce the inductance of the transformer from 10 H to a reasonable value,,, say 200 mh. You immediately see that the phase shift due to winding capacitance disappear. And if that does not solve your problem,. Leave another comment.

kiani

7 years ago

Plus if the 2 ohm and 4 ohms represent the winding resistance, then the ( 8) ohm resistor is connected to the wrong end of the 4 ohms.

add comment in editor

EveryCircuit is an easy to use, highly interactive circuit simulator and schematic capture tool. Real-time circuit simulation, interactivity, and dynamic visualization make it a must have application for professionals and academia. EveryCircuit user community has collaboratively created the largest searchable library of circuit designs. EveryCircuit app runs online in popular browsers and on mobile phones and tablets, enabling you to capture design ideas and learn electronics on the go.