vb.net barcode reader code ~ I1 + ~ V1 _ Primary n1:n2 or 1:N ~ I2 + ~ V2 _ Secondary in Software Drawer Quick Response Code in Software ~ I1 + ~ V1 _ Primary n1:n2 or 1:N ~ I2 + ~ V2 _ Secondary

~ I1 + ~ V1 _ Primary n1:n2 or 1:N ~ I2 + ~ V2 _ Secondary
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Figure 738 Ideal transformer
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An ideal transformer multiplies a sinusoidal input voltage by a factor of N and divides a sinusoidal input current by a factor of N
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If N is greater than 1, the output voltage is greater than the input voltage and the transformer is called a step-up transformer If N is less than 1, then the transformer is called a step-down transformer, since V2 is now smaller than V1 An ideal transformer can be used in either direction (ie, either of its coils may be viewed as the input side or primary) Finally, a transformer with N = 1 is called an isolation transformer and may perform a very useful function if one needs to electrically isolate two circuits from each other; note that any DC currents at the primary will not appear at the secondary coil An important property of ideal transformers is conservation of power; one can easily verify that an ideal transformer conserves power, since 1 2 S1 = I V1 = N I V2 2 = I V2 = S2 N (734)
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That is, the power on the primary side equals that on the secondary
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~ I1 + ~ V1 _ + ~ V2 _ n2 ~ ~ V2 = V n1 1 n3 ~ ~ V3 = V n1 1
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+ n3 ~ V3 _
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Figure 739 Center-tapped transformer
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In many practical circuits, the secondary is tapped at two different points, giving rise to two separate output circuits, as shown in Figure 739 The most common con guration is the center-tapped transformer, which splits the secondary voltage into two equal voltages The most common occurrence of this type of transformer is found at the entry of a power line into a household, where a high-voltage primary (see Figure 764) is transformed to 240 V, and split into two 120-V lines Thus, V2 and V3 in Figure 739 are both 120-V lines, and a 240-V 2 + V3 ) is also available line (V
EXAMPLE 712 Ideal Transformer Turns Ratio
Problem
We require a transformer to deliver 500 mA at 24 V from a 120-V rms line source How many turns are required in the secondary What is the primary current
Solution
Known Quantities: Primary and secondary voltages; secondary current Number of turns in the primary coil
Find: n2 and I1 500 mA; n1 = 3,000 turns Schematics, Diagrams, Circuits, and Given Data: V1 = 120 V; V2 = 24 V; I2 =
Assumptions: Use rms values for all phasor quantities in the problem Analysis: Using equation 733 we compute the number of turns in the secondary coil as
follows: V2 V1 = n1 n2 n2 = n 1 V2 24 = 600 turns = 3,000 120 V1
Knowing the number of turns, we can now compute the primary current, also from equation 733: n 1 I1 = n 2 I 2 n2 600 I1 = I2 = 500 = 100 mA n1 3,000
Comments: Note that since the transformer does not affect the phase of the voltages and
currents, we could solve the problem using simply the rms amplitudes
EXAMPLE 713 Center-Tapped Transformer
Problem
A center-tapped power transformer has a primary voltage of 4,800 V and two 120-V secondaries (see Figure 739) Three loads (all resistive, ie, with unity power factor) are connected to the transformer The rst load, R1 , is connected across the 240-V line (the two outside taps in Figure 739) The second and third loads, R2 and R3 , are connected across each of the 120-V lines Compute the current in the primary if the power absorbed by the three loads is known
Part I
Circuits
Solution
Known Quantities: Primary and secondary voltages; load power ratings
Find: Iprimary Schematics, Diagrams, Circuits, and Given Data: V1 = 4,800 V; V2 = 120 V; V3 = 120 V P1 = 5,000 W; P2 = 1,000 W; P3 = 1,500 W
Assumptions: Use rms values for all phasor quantities in the problem Analysis: Since we have no information about the number of windings, nor about the
secondary current, we cannot solve this problem using equation 733 An alternative approach is to apply conservation of power (equation 734) Since the loads all have unity power factor, the voltages and currents will all be in phase, and we can use the rms amplitudes in our calculations: Sprimary = Ssecondary or Vprimary Iprimary = Psecondary = P1 + P2 + P3 Thus, 4,800 Iprimary = 5,000 + 1,000 + 1,500 = 7,500 W 7,500 W = 15625 A Iprimary = 4,800 A
Impedance Re ection and Power Transfer As stated in the preceding paragraphs, transformers are commonly used to couple one AC circuit to another A very common and rather general situation is that depicted in Figure 740, where an AC source, represented by its Th venin equivalent, e is connected to an equivalent load impedance by means of a transformer It should be apparent that expressing the circuit in phasor form does not alter the basic properties of the ideal transformer, as illustrated in the following equation: V2 V1 = N I1 = N I2