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enter image description here

In the above figure (step up transformer), We can calculate \$V_s\$ using \$\frac{V_s}{V_p}=\frac{N_s}{N_p}\$ and \$V_s=50 V\$

So, power dissipated in the resistance, \$P=\frac{V_s^2}{R}=2500 W\$ (which is clearly impossible as I delivered only \$P=V_pI_p=5W\$ from principal coil.)

Again,if I calculate power dissipated in \$R\$ using \$I_s(=0.1 A\$ using \$\frac{I_s}{I_p}=\frac{N_p}{N_s}\$) , \$P=I_s^2R=0.01W.\$

Powers calculated in both ways are supposed to be same, aren't they??

I think, my main problem is I can't find out the current passing through \$R\$ as \$\frac{I_s}{I_p}=\frac{N_p}{N_s}\$ gives \$0.1 A\$ whereas \$i=\frac{V_s}{R}\$(Ohm's law) gives \$50A\$.

What am I missing here?Any help would be highly appreciated.

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  • \$\begingroup\$ The primary current is determined by the secondary current, which in turn is determined by the load and secondary voltage. \$\endgroup\$
    – Chu
    May 31 '16 at 22:58
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You can't constrain the primary voltage and current at the same time.

It's no different from saying I have a 10mOhm resistor with 5V across it and 1A through it. It violates Ohm's law.

The reflected impedance of the 1 ohm resistor on the primary side is 1/(10)^2 or 10mOhm. So with 5V across 10mOhm you'll have 500A primary current. On the other hand if you source 1A in the primary the primary voltage will be 10mV.

If you want to meet the primary constraints of 5V and 1A you have you need a reflected impedance of 5 ohms in the primary, meaning a 500 ohm resistor on the secondary.

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  • \$\begingroup\$ okay, but in the figure of the following link, one can find a value for Rp so that Vp= 5 V and Ip= 1 A, can't they?? I am not familiar with the calculation of reflected impedance, and it seemed quite complex to me at some quick searches, so I couldn't find the exact value of Rp for this case. imgur.com/KynzFvw \$\endgroup\$ May 31 '16 at 21:28
  • \$\begingroup\$ Sure, Rp is 5 ohms minus 10mOhm or 4.99 ohms. If you work through the characteristics of an ideal transformer the reflected impedance easily falls out. Impedance is reflected as the turns ratio squared. \$\endgroup\$
    – John D
    May 31 '16 at 21:52
  • \$\begingroup\$ So, it seems that we can constrain both primary current and voltage at the same time just by making good choice of R, doesn't it?? Then what's the problem here?? \$\endgroup\$ May 31 '16 at 21:58
  • \$\begingroup\$ In fact you can not. The voltage across the primary is not 5V once you add the series resistor. The voltage across the primary is only 10mV. The voltage across the series resistor is 4.99V, but that voltage doesn't appear across the primary of the transformer. If you are talking about a non-ideal transformer with some winding resistance then the model is a series resistor with an ideal transformer (and maybe a bunch of other stuff) but it's analyzed the same way. \$\endgroup\$
    – John D
    May 31 '16 at 22:30
  • \$\begingroup\$ You don't state whether the transformer in the original question is ideal or not, and no winding resistances or leakage or magnetizing inductances are given so it's logical to assume it's an ideal transformer. However, if the transformer wasn't intended to be ideal, the problem could have been stated as "what primary winding resistance would be required to satisfy the given conditions". That would assume that the secondary resistance + load = 1 ohm and that leakage inductances were negligible at the operating frequency, etc. \$\endgroup\$
    – John D
    May 31 '16 at 22:42
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A transformer passes power (voltage times current), not voltage and current independently.

If I plug my 1200 watt kettle into a 120 volt outlet, it will draw 10 Amps.

If the local HV power distribution is 12 KV (12,000 volts) the transformer on the power pole will have a 100:1 turns ratio, to step the voltage down from 12KV to the 120 volts I want. That 100:1 turns ratio will step the 10 Amps my kettle draws down to 0.1 Amp at 12,000 volts - that's still 1200 watts.

(For the pedants: I've neglected losses in the transformer - it will actually draw a little more than 1200 watts, allowing for losses. Also, in North America, the transformer secondary really delivers 120/240 volts from a centre-tapped winding.)

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