Center Tapped Transformer

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Strings (SiPjAjk) = S7P3A32     Base Sequence = 12735     String Sequence = 12735 - 3 - 32



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Center Tapped Transformer
Figure 8.42 shows a center-tapped Transformer. The following information is given about the transformer:
Voltages and current are rms values.
Primary voltage = 4,800 V
Secondary voltage of 240 V is split (because transformer is center tap) into two voltages:
V2 = 120 V; V3 = 120 V.
Three resistive loads (R1, R2, R3) are connected to the transformer (connection not shown in figure 8.42).
R1 is connected to the 240 V line.
R2 and R3 are connected to each of the 120 V lines.
Determine the power absorbed by each of the loads, if:
Power absorbed by R2 = P2
Power absorbed by R1 = 5P2
Power absorbed by R3 = 1.5P2
Ccurrent through primary coil, I1 = 1.5625 A.

The string:
S7P3A32 (Force - Push).
The math:
Center Tapped Transformer
Pj Problem of interest is of type force. The transformer steps up or steps down voltage. Voltage, power and energy problems are force problems (force-push).

The loads are all resistive, hence their respective power factor = 1.
So, voltages and currents are in phase.
Therefore rms amplitudes can be used in the calculations:
|Sprimary| = |Ssecondary| = 4,800 x 1.5625 = 7,500.
So, 5P2 + P2 + 1.5P2 = 7,500
P2 = 1,000.
Hence, power absorbed by the resistors are as follows:
Powere absorbed by R1 = 5,000 W
Power absorbed by R2 = 1,000 W
Power absorbed by R3 = 1,500 W.

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