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I have researched a lot on this topic (including this site) and found this link somewhat helpful : How is it possible to have high voltage and low current? ...     But still it seems to fall somewhat short.

People have explained that since power must remain constant(no losses in ideal case) then power in = power out, i.e.

$$V_{in} \cdot I_{in} = V_{out} \cdot I_{out}$$

and

$$\dfrac{V_{in}}{V_{out}} = \text{Ratio of number of turns in primary to that in secondary coil}$$

But how is this possible? Let's say the plant can supply power of 1000 W, @1000 V and 1 Amp. The voltage is determined by the generator but the current is determined by the impedance of the circuit having the primary coil. Lets say the transformer is step down and the voltage induced (RMS) is 200V. According to previous logic, the current must be 5 Amps. But shouldn't it depend on the resistance(or impedance) of the appliance we have connected to the secondary coil?

EDIT 1 Why does the primary impedance depend on Np/Ns Ratio? Is it due to mutual induction? And why does actual load impedance in secondary depend on that ratio and not what appliances we have connected across the secondary coil? If I connect different number of appliances will the voltage induced across the secondary change?

Edit 2 If the primary impedance is changed by connecting /disconnecting "appliances" then will the apparent power to be supplied by generator change? What about actual power? (Although power in = out, I'm talking about the value) If so, how does the generator manage to change the power supply?

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  • \$\begingroup\$ Could it be that (in ideal case) if nothing is connected to the secondary coil, the impedance of the first coil is then zero, and thus, the primary current and power is zero? \$\endgroup\$ – Nazar Sep 9 '15 at 18:41
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    \$\begingroup\$ You have stepped down 5:1 so the current must be 5 amps. The trick is, what you connect to the secondary is a new appliance, with a much lower load resistance (impedance) than the original : n^2 lower, or (for n=5) 1/25 of the original impedance. Instead of the original 1000 ohm load, its impedance is 200/5 = 40 ohms. \$\endgroup\$ – Brian Drummond Sep 9 '15 at 19:09
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You're right, the current does depend on the impedance connected to the secondary coil. I think you're getting confused about what rules to follow when. Here are the rules that always apply to ideal transformers:

$$\frac {V_P}{V_S} = \frac {N_P}{N_S}$$

$$\frac {I_P}{I_S} = \frac {N_S}{N_P}$$

$$\frac {Z_P}{Z_S} = \left( \frac {N_P}{N_S} \right)^2$$

$$P_P = P_S$$

Just because your power source can supply 1000 volts at 1 amp doesn't mean it will under all circumstances. What we can say is that if the source is supplying 1000 V and 1 A into the primary side of an ideal transformer, then:

  • The current on the secondary side is \$1\cdot\frac {N_P}{N_S}\$ amps.
  • The voltage on the secondary side is \$1000 \cdot\frac {N_S}{N_P}\$ volts.
  • The apparent input impedance on the primary side has a magnitude of 1000 ohms.
  • The actual load impedance on the secondary side has a magnitude of \$1000\left(\frac {N_S}{N_P}\right)^2\$ ohms.
  • The input power on the primary side and the output power on the secondary side are both 1000 watts.

If you change the turns ratio \$\frac{N_P}{N_S}\$, the apparent impedance of the primary side will change, just as if you had changed the actual load impedance. The voltages and currents on both sides will change accordingly.

EDIT: Yes, the primary impedance (also known as the reflected impedance) depends on the turns ratio due to mutual inductance. The actual load impedance on the secondary does not -- it's the physical impedance connected across the wires, what you're calling "appliances". But just as with Ohm's Law, if you know the voltage and current on the primary side and the turns ratio, you can calculate the impedance on the secondary side. That's what I'm doing above.

To put it another way: the voltage applied to the primary side determines the voltage seen on the secondary side. The secondary voltage and the load impedance determine the secondary current. The secondary current determines the primary current. The primary voltage and primary current give you the reflected (primary) impedance.

Think of the reflected impedance as a Thevenin equivalent. If you connect a 22k resistor to the secondary side of a transformer where \$\frac {N_P}{N_S} = 10\$, then the primary side of the transformer acts just like a 220 ohm resistor.

See also this question.

EDIT 2: I'm not an expert on generators, but I'll try to answer your question anyway. :-)

As a rule of thumb, the voltage of a generator depends on its speed. Turning the generator at a constant speed produces a constant voltage (an AC voltage, in this case). When applied to a load, this voltage causes current to flow. The current acts as an electromagnet and puts a torque on the generator, opposing its motion. Overcoming that torque consumes mechanical energy. The mechanical energy consumed is equal to the electrical energy produced. (I'm ignoring friction and inertia, and only talking about the steady state.)

If you feed a constant amount of mechanical power into the generator, its voltage will vary based on the load resistance, keeping the electrical power equal to the mechanical power. But in practice, we normally want a generator to produce a constant voltage. So the mechanical power is varied to maintain the voltage.

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  • \$\begingroup\$ Why does the primary impedance depend on Np/Ns Ratio? Is it due to mutual induction? And why does (according to your point 4) actual load impedance in secondary depend on that ratio and not what appliances we have connected across the secondary coil? If I connect different number of appliances will the voltage induced across the secondary change? \$\endgroup\$ – frags51 Sep 10 '15 at 8:26
  • \$\begingroup\$ I've updated my answer. \$\endgroup\$ – Adam Haun Sep 10 '15 at 13:53
  • \$\begingroup\$ One more doubt.. if the primary impedance is changed by connecting /disconnecting "appliances" then will the apparant power to be supplied by generator change? What about actual power? (Although power in = out, I'm talking about the value) If so, how does the generator manage to do so? \$\endgroup\$ – frags51 Sep 10 '15 at 14:37
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    \$\begingroup\$ Yes and yes. That's exactly what happens when you plug something into the wall. Can you clarify your question about the generator? \$\endgroup\$ – Adam Haun Sep 10 '15 at 15:32
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    \$\begingroup\$ I updated my answer again. \$\endgroup\$ – Adam Haun Sep 10 '15 at 16:27
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A transformer, what its name suggests transforms different voltage levels. You can use an equivalent circuit to find out that the current in the primary is proportional to the load applied on secondary. If the secondary is left open, then only magnetizing current flows trough primary winding (very little current, 90 degress out of phase).

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  • \$\begingroup\$ can you give the equivalent circuit diagram? \$\endgroup\$ – frags51 Sep 10 '15 at 11:24
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How is it possible to have high voltage and low current? This is possible if Power, P, is held constant.

P=VI ==> V=P/I ...Therefore V is inversely proportional to I

with V=P/I, If P is held constant, an increase in V will lead to a decrease in I

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For years the common explanation never quite satisfied me. It finally dawned on me that:

$$ \frac{V_p}{V_n} = \frac{N_p}{N_n} $$

is a consequence and that

$$ N_p * I_p = N_s * I_s $$

is the real driving factor. That is, in an ideal transformer the amp-turns in the primary are equal to the amp-turns in the secondary. In a real transformer you usually just have to add some leakage to the primary and you have a very accurate model.

Now think about the consequences of this driving factor - the power supply connected to the primary winding is at some voltage and will push a certain amount of current through the winding. Just like in Ohm's law, the current is determined by the impedance of the winding. The winding's impedance is determined (mostly) by the reflected impedance of the secondary. The primary presents the reflected impedance and the resulting primary current flows. The transformer maintains amp-turns and so a resulting current flows in the secondary. When you finally do the voltage calculation you find that the Vp/Vn ratio holds, but really only as a consequence. Sometimes the Vp/Vn ratio does not hold and working through it this way will generally reveal why.

Since I've flipped my reasoning about transformers to this amp-turns dominant mode, a great deal of the consequences fall into line much more intuitively.

For example, to answer your original question, the primary 1000V is available to the reflected load. The actual current (and therefore power) will be determined by that load.

To answer your Edit 1: because the transformer must push a certain number of amp-turns, the secondary impedance appears as a reflected impedance on the primary.

To answer your Edit 2: Yes, the power changes. Start with the impedance of the load, calculate the reflected impedance, calculate the resulting primary current and then you have the delivered power. If you add another load the impedance goes down, the current goes up and therefore the power goes up. A transformer's power rating is a maximum, not a constant.

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