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There is a transformer with \$600V/120V\$, and the impedance in the high voltage side is \$52-30j Ω\$, and there is a loading, which impedance is \$0.8∠10°Ω\$, connected with the low voltage side. Now when the \$V_2=120∠0°V\$, what is the value of the current in the primary side??

My thinking

\$V=IZ,\$ so \$I_1=\frac{V_1}{Z_1}\$, that is, current in the primary side \$I_1=\frac{600}{52-30j}=\frac{600}{60∠\theta }=10∠-\theta A\$

Solution

Current in the secondary side \$I_2=\frac{V_2}{Z_2}=\frac{120∠0°}{0.8∠10°}=150∠-10°A\$,

so current in the primary side \$I_1=I_2 \times \frac{N_2}{N_1}=I_2 \times \frac{V_2}{V_1}=150∠-10° \times \frac{120}{600}=30∠-10°A\$

The solution just use \$I=\frac{V}{z}\$ to calculate the current in the secondary side,\$I_2\$, then use the ratio between voltage and current to calculate the current in the primary side, \$I_1\$.

However, we have the voltage and impedance in the primary side, why don't we just use \$I_1=\frac{V_1}{Z_1}\$ to calculate the current in the primary side?Why will we get the wrong answer if we use \$I_1=\frac{V_1}{Z_1}\$ to calculate \$I_1\$ directly?

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  • \$\begingroup\$ In "take enough steps back to see the big picture" terms, because if you did so you would be ignoring the load. \$\endgroup\$ – Chris Stratton May 21 '20 at 15:26
  • \$\begingroup\$ You haven't drawn the circuit but how do you know that the impedance on the primary side is across the input voltage 600V? I=V/Z holds true if that 52-j30 Ohm impedance were in parallel to the input souce. And if that is the case, the current through that impedance, still would not be the primary current going into the primary side of the transfomer. \$\endgroup\$ – Big6 May 21 '20 at 16:26
  • \$\begingroup\$ How can 52 - j30 suddenly become 60 angle zero? the angle should be -30 degrees. Draw a circuit and mark on it the various currents (I1 and I2) and impedances like Z1. \$\endgroup\$ – Andy aka May 21 '20 at 17:05
  • \$\begingroup\$ @Andyaka 60 angle "\$\theta\$",not "0" \$\endgroup\$ – XM551 May 21 '20 at 23:46
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You haven't drawn the circuit but how do you know that the impedance on the primary side is across the input voltage 600V?

I=V/Z holds true if that 52-j30 Ohm impedance were in parallel to the input souce. And if that is the case, the current through that impedance, still would not be the primary current going into the primary side of the transfomer.

This is what you're actually doing:

schematic

simulate this circuit – Schematic created using CircuitLab

\$I_1\$ is the current through the 50-j30 impedance, but is not the one that actually flows into the primary side.

Maybe, just maybe this is what you really have:

schematic

simulate this circuit

And in this case, all the primary current flows through the 50-j30 impedance, and it makes sense to find that current with the \$I_P\cdot N_1=I_S\cdot N_2\$ relationship since you already have all you need.

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  • \$\begingroup\$ Thanks a lot,now i know the reason now \$\endgroup\$ – XM551 May 22 '20 at 0:07

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