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An AC analog amp-meter with a range of 0 to 1000 amps is placed in the primary of the circuit to determine the primary current as shown below. If both transformers are subtractive polarity, what would the amp meter read?
Answer to nearest hundredth.
Answer
Since we’re assuming ideal transformers (transformers a & b):
I_pri = I_pri_a = I_pri_b
I_sec = I_sec_a = I_sec_b
4 * I_pri_a = I_sec_a
3 * I_pri_b = I_sec_b
The only number that satisfies all of these equations is I_pri = 0.
MMF balance is impossible. The network draws no current and the input impedance is infinity (or the same as an open circuit). It is completely irrelevant what is connected on the secondary.
Another way to think about this- If both primaries where connected in parallel we would have a short circuit. When you connect them in series you have an open circuit.
1. Solving for equivalent impedance of the parallel load branch results in total load impedance magnitude of 50.225ohms.
2. Since both transformers are connected with subtractive polarity. The primary has 7 (4+3) effective Coil turns and the secondary has 2. The effective turns ratio of circuit becomes 7:2.
3. Secondary voltage is (2/7)(240V) = 68.57V
Secondary current magnitude = 68.57V / 50.225 ohms = 1.365A
4. Primary current magnitude read by ammeter = (1.265A)(2/7) = 0.39A
This is incorrect.
This seems to be a dual primary and dual secondary connected in series.
Vsource = 240 Vrms
The two windings are not identical so we need to establish coil/voltage ratio. We know that the sum Va @N=4 + Vb @N=3 = 240 V. Using auto transformer relationship. Vb/Va = Nb/Na, and replacing Va for 240 – Vb, we can see that Vap= 137.143 and Vbp= 102.857. Once we have the two primary voltages, we can use the transformer ratio formulas to get all the secondary voltages. Notice that Vas=34.286 and Vbs=34.286. Finally, knowing that Vs= Is*Zs, we find Zs=50.1678 (magnitude), which is connected across both series windings. Adding the voltages to Vtotalsec= 68.562, we see that Itotsec= 1.367 A. Using ratio formulas, Vptot/Vstot=Istot/Iptot, the current on the primary side read by the ammeter should be 2.56A.
This is incorrect.
It seems like an assumption needs to be made to reach a solution. Assuming ideal transformers and no losses, if for simplicity, 10 A is assumed flowing from the source through the two primaries in series, then by transformer action, 40A is flowing in the secondary of the upper transformer and 30 A is flowing in the secondary of lower transformer. The cannot be true as they are the same current as they are in series.
When questions like this are on the PE- it is assumed that the transformer are ideal.
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