Question
0. A power plant produces 1200 MW of power that is sent through a transmission line with a resistance of (2)/(3)Omega . A transformer is used that has 120 windings I in the primary circuit and 600 windings in the secondary circuit. If this changes the power loss to 0.20% what are the current and potential difference both before and after the transformer change? Assume there is no power loss in the transformer. (13.5,13.6) T/I
Solution
4.6
(212 Votes)
Graham
Elite · Tutor for 8 years
Answer
Here's how to solve this problem:**Understanding the Problem**We're dealing with power transmission and the effect of a step-up transformer on reducing power loss due to line resistance. The transformer increases the voltage and decreases the current, minimizing power loss (which is proportional to the square of the current).**Formulas*** **Power:** P = IV (Power = Current x Voltage)* **Power Loss:** P_loss = I²R (Power loss = Current squared x Resistance)* **Transformer Voltage Ratio:** V₂/V₁ = N₂/N₁ (Secondary voltage / Primary voltage = Number of secondary windings / Number of primary windings)* **Transformer Current Ratio:** I₁/I₂ = N₂/N₁ (Primary current / Secondary current = Number of secondary windings / Number of primary windings)**Calculations**1. **Initial Conditions (Before Transformer):*** Power Loss Percentage: Let's assume the initial power loss is 'x'. We don't need to calculate 'x' directly yet.* Power: P = 1200 MW = 1.2 x 10⁹ W* Resistance: R = 2/3 Ω2. **Final Conditions (After Transformer):*** Power Loss Percentage: 0.20% = 0.002* Power: P = 1200 MW (Power remains the same, assuming ideal transformer)* Resistance: R = 2/3 Ω (Resistance of the transmission line is unchanged)* Power Loss: P_loss = 0.002 * 1.2 x 10⁹ W = 2.4 x 10⁶ W3. **Current After Transformer:*** P_loss = I₂²R* I₂ = sqrt(P_loss / R) = sqrt(2.4 x 10⁶ W / (2/3 Ω)) = 1897.37 A ≈ 1900 A4. **Voltage After Transformer:*** P = I₂V₂* V₂ = P / I₂ = (1.2 x 10⁹ W) / 1897.37 A ≈ 632455.53 V ≈ 632 kV5. **Current Before Transformer:*** I₁/I₂ = N₂/N₁* I₁ = I₂ * (N₂/N₁) = 1897.37 A * (600/120) = 9486.85 A ≈ 9500 A6. **Voltage Before Transformer:*** V₁/V₂ = N₁/N₂* V₁ = V₂ * (N₁/N₂) = 632455.53 V * (120/600) = 126491.11 V ≈ 126 kV**Results*** **Before Transformer:** Current (I₁) ≈ 9500 A, Voltage (V₁) ≈ 126 kV* **After Transformer:** Current (I₂) ≈ 1900 A, Voltage (V₂) ≈ 632 kV**Important Note:** The textbook answers (13.5, 13.6) are likely incorrect. The calculations above demonstrate the correct method and provide reasonable values for the current and voltage before and after the transformer. The significant reduction in current after the step-up transformer is what leads to the much lower power loss.