The Ideal Transformer Will Not Be Treated During Lecture But

The Ideal Transformer Will Not Be Treated During Lecture But The T The ideal transformer will not be treated during lecture, but the textbook includes a section that covers the topic well. Read that section. Be sure that you are reading the section on the ideal transformer. This material is testable.

Paper For Above instruction The problem set revolves around understanding the ideal transformer’s behavior and modeling, analyzing scenarios with load and short circuit, and calculating power losses, efficiencies, and system characteristics under various conditions. We begin with fundamental principles, advancing to practical implications in industrial contexts and the limitations of idealized models.

Part 2(a): Transformer Voltage Transformation Given a transformer with a primary-to-secondary turns ratio of 100:1, and an applied primary voltage of 27.7 kV (rms) at 60 Hz, the secondary voltage for an ideal transformer can be calculated directly from the turns ratio. The voltage transformation ratio equals the turns ratio, so: V_secondary = V_primary / turns_ratio = 27,700 V / 100 = 277 V (rms).

This implies that, without any load considerations, the transformer steps down the voltage from 27.7 kV to 277 V across its secondary winding. The diagram illustrates the primary coil connected to the source and the secondary load, with voltage levels indicated accordingly.

Part 2(b): Secondary Current with a 2-Ω Load The secondary load is 2.00 Ω. Using Ohm’s law, the current drawn from the secondary is: I_secondary = V_secondary / R_load = 277 V / 2 Ω = 138.5 A. This high current reflects the nature of the step-down transformer reducing voltage, which consequently increases current proportionally, assuming an ideal scenario.

Part 2(c): Primary Current Calculation For an ideal transformer, power on the primary equals power on the secondary, neglecting losses: P_secondary = V_secondary × I_secondary = 277 V × 138.5 A ≈ 38,345 W. The primary voltage is 27.7 kV, so primary current is:

I_primary = P_secondary / V_primary = 38,345 W / 27,700 V ≈ 1.38 A. This illustrates the current transformation ratio, where the primary current is substantially smaller than the secondary current due to voltage reduction.

Part 2(d): Power Transfer and Conservation The power delivered to the load (secondary side) is approximately 38.3 kW. Due to the ideal transformer assumption, the primary must draw the same power, also about 38.3 kW . This is consistent with the first law of thermodynamics — no energy is lost in an ideal transformer. The primary power is thus equal to the secondary power, confirming energy conservation.

Part 2(e): Consistency of Power Calculations Since the primary power and secondary power are equal (~38.3 kW), and the primary current (~1.38 A) and secondary current (138.5 A) satisfy the power equation, the calculations are consistent. Specifically: Power_primary = V_primary × I_primary ≈ 27,700 V × 1.38 A ≈ 38,226 W, very close to the secondary power, considering rounding errors.

The slight discrepancy accounts for rounding differences, but overall, the calculations uphold energy conservation in an ideal scenario.

Part 2(f): Impedance as Seen by Utility Using the secondary load impedance of R = 2 Ω and the turns ratio of 100:1, the reflected impedance seen at the primary is: Z_primary = R_load × (turns ratio)^2 = 2 Ω × 100^2 = 2 Ω × 10,000 = 20,000 Ω = 20.0 kΩ.

Thus, when a 2.0-Ω load is connected across the secondary, the utility “sees” a 20.0 kΩ resistor.

Part 2(g): Open-Circuit Scenario Description If the secondary is disconnected (open circuit), no current flows in the secondary, and the transformer’s primary side appears as a very high impedance, effectively an open circuit from the utility’s perspective.

Due to the high turns ratio, the reflected impedance becomes very large, so the primary current remains minimal, similar to an open circuit. However, in real, non-ideal transformers, some small magnetizing

current exists even when secondary is open, but this is typically negligible and limited by design features.

Part 2(h): Short Circuit and Equivalent Inductive Behavior In the short-circuit condition, the secondary is directly connected across the transformer. The primary appears as an inductor with inductance L = 0.5 H. The rms current in the primary under short circuit conditions is calculated by: I_primary_short = V_primary / (jωL) = 27,700 V / (j × 2π × 60 Hz × 0.5 H) ≈ 27,700 V / j(188.5) ≈ 147 A RMS. The magnitude of the impedance is Z = ωL, which is approximately 188.5 Ω, leading to a high but finite current limited by the inductance.

Part 2(j): Time Domain Expressions for Voltage and Current The voltage across the primary in the short circuit is V(t) = V_peak × sin(ωt), where V_peak is the rms value times √2: V_peak = 27,700 V × √2 ≈ 39,173 V. V(t) = 39,173 V × sin(ωt).

Similarly, the primary current is: i(t) = I_peak × sin(ωt + φ), where I_peak = √2 × I_rms ≈ 147 A, and φ is the phase angle depending on inductive reactance. Since the impedance is mainly inductive, φ ≈ 90°, meaning current lags voltage by 90°: i(t) ≈ 147 A × sin(ωt - 90°) = -147 A × cos(ωt).

Part 2(k): Circuit Breaker and Short-Circuit Duration The circuit breaker will interrupt the short-circuit current after a brief moment, limited only by the inductance and resistance in the real system. In practice, the current lasts only milliseconds, preventing catastrophic damage. The inductor’s energy temporarily sustains the current shortly after the breaker opens, but the breaker’s operation rapidly reduces the current to zero.

Part 2(m): Power Loss in Utility Wires Each wire has a resistance of 1.50 Ω/mile, so over 50 miles:

R_wire = 1.50 Ω/mile × 50 miles = 75 Ω per wire.

Since there are two wires, total resistance is: R_total = 2 × 75 Ω = 150 Ω.

The current flowing through these wires (from part c) is 1.38 A. Power loss due to resistance is: P_loss = I^2 × R_total = (1.38 A)^2 × 150 Ω ≈ 285 W. Part 2(n): System Efficiency Efficiency η is the ratio of useful power delivered to the load over total power supplied including losses: η = P_load / (P_load + P_loss) = 38,345 W / (38,345 W + 285 W) ≈ 0.9926 or 99.26%.

Part 2(p): No-Transformer Scenario If no transformer were used, the power would be directly supplied at 277 V across the 2 Ω load: P_direct = V^2 / R = (277 V)^2 / 2 Ω ≈ 38,435 W. Similarly, the power loss in the transmission wires remains approximately 285 W, and the efficiency is almost the same, but the voltage at the plant’s end would need to be the same as the supply, requiring thicker, more costly conductors and higher voltage handling capabilities along the entire line.

Part 2(q): Why Not Use 277 V Utility Voltage Directly The utility does not supply power at 277 V directly because higher voltages are preferable for long-distance power transmission due to reduced current and minimal losses, which is consistent with the earlier calculations. Using a higher voltage (like 27.7 kV) significantly reduces the current for the same power level, drastically lowering resistive losses. Additionally, the safety standards and insulation requirements at 277 V within the plant are manageable, but transmitting at high voltage outdoors from substations is safer and more economical.

Part 3: Power at 27.7 kV DC If the power source supply is DC at 27.7 kV, the transformer’s primary will not function because transformers rely on changing magnetic flux—generated by alternating current—to induce voltage in the secondary. For DC, the magnetic flux in the core remains constant, resulting in zero induced voltage; thus, no power is transferred to the secondary. So, the load receives no power in this case, illustrating the

fundamental necessity of AC for transformer operation.

In conclusion, the series of analyses demonstrate key principles of transformer operation, impedance reflections, power losses, efficiency, and practical limitations within power systems, emphasizing the importance of AC and appropriate voltage levels in electrical transmission and distribution.

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