Let's go through the problem step-step. --- # Given Data: Transformer turns ratio 1:5 (primary:) - Voltage source: \( \,kV) on primary- Resistor on secondary: \( 30\,\ \) - Power base: \(100\,MVA \) --- Part (a): Calculate load power and source without per-unit system ### Step 1: Find the secondary voltage \( V_{b} \) Since the transformer is 1:5, the secondary voltage: \[ V_{b} = \frac{V_{primary}}{5} = \frac{25\,kV}{5} = 5\,kV \] ### Step 2: Calculate the secondary current \( I_{b} \) Using the resistor \( R_b = 30\,\Omega \): \[ I_b = \frac{V_b}{R_b} = \frac{5\,kV}{30\,\Omega} = \frac{500\,V}{30\,\Omega} \approx 166.67\,A \] ### Step 3: Calculate the secondary power \( P_b \) \[ P_b = V_b \times I_b = 500\,V \times 166.67\,A = 833,333\,W \approx .833\,MW \] ### Step 4: Determine the load power in MVA Since the power is primarily resistive, the apparent power: \[ S_b = V_b \times I_b = .833\,MW \] Expressed in MVA: \[ \boxed{ \text{Load Power} = .833\,\text{MVA} } \] ### Step 5: Calculate the primary source current \( I_{primary} \) The transformer power transfer is: \[ S_{primary} = S_b = .833\,MVA \] The primary voltage: \[ V_{primary} = 25\,kV \] So, the primary current: \[ I_{primary} = \frac{S_{primary}}{V_{primary}} = \frac{.833\,MVA}{25\,kV} = \frac{833,333\,VA}{25,000\,V} \approx 33.33\,A \] --- # **Summary for Part (a):** | Quantity | Value | |---|---| | Load Power \( S_b \) | **.833 MVA** | | Source Current \( I_{primary} \) | **33.33 A** | --- ## Part (b): Create per-unit base table ### Step 1: Define base quantities - Power base: \( S_{base} = 100\,MVA \) ### Step 2: Calculate per-unit values for the secondary side - \( S_b = .833\,MVA \) - \( V_b = 5\,kV \) - \( I_b = 166.67\,A \) - \( Z_b = \frac{V_b^2}{S_b} \) \[ Z_b = \frac{(500\,V)^2}{.833 \times 10^6\,VA} = \frac{25 \times 10^6}{.833 \times 10^6} \approx 30\,\Omega \] ### Step 3: Primary side per-unit values - \( S_{primary} = .833\,MVA \) - \( V_{primary} = 25\,kV \) - \( I_{primary} = 33.33\,A \) - \( Z_{primary} = \frac{V_{primary}^2}{S_{primary}} \) \[ Z_{primary} = \frac{(25,000\,V)^2}{.833 \times 10^6\,VA} = \frac{625 \times 10^6}{833,333} \approx 750\,\Omega \] ### **Per-unit values:** | Quantity | Primary | Secondary | |---|---|---| | \( S_b \) | \( \frac{.833\,MVA}{100\,MVA} = .00833 \) PU | \( .833\,MVA \) (given) | | \( V_b \) | \( \frac{25\,kV}{25\,kV} = 1. \) PU | \( \frac{5\,kV}{25\,kV} = .2 \) PU | | \( I_b \) | \( \frac{33.33\,A}{100\,000\,A} = .0003333 \) PU | \( \frac{166.67\,A}{100\,000\,A} = .0016667 \) PU | | \( Z_b \) | \( \frac{30\,\Omega}{750\,\Omega} = .04 \) PU | \( 30\,\Omega \) (calculated) | --- ## Part (c): Calculate load power and source current using per-unit system ### Step 1: Convert resistor to per-unit \[ Z_{b} = 30\,\Omega \quad (\text{already calculated as } .04\,PU) \] ### Step 2: Load power in per-unit \[ S_b = .00833\,PU \times 100\,MVA = .833\,MVA \] ### Step 3: Calculate the secondary current \( I_b \) in per-unit \[ I_b = \frac{V_b}{Z_b} = \frac{.2\,PU}{.04\,PU} = 5\,PU \] But, note that in actual calculations, power is: \[ P_b = V_b \times I_b \times \cos{\phi} \] Since power factor isn't specified, assume resistive load (\(\cos{\phi} \approx 1\)): \[ S_b = V_b \times I_b \Rightarrow I_b = \frac{S_b}{V_b} = .833\,MVA / 5\,kV = 166.67\,A \] which matches previous calculations. ### **Final answers:** | Quantity | Load Power | Source Current | |---|---|---| | **Load power** | **.833 MVA** | — | | **Source current** | — | **33.33 A** | --- # **Summary:** - **Part (a):** Load power ≈ .833 MVA, source current ≈ 33.33 A. - **Part (b):** Per-unit values tabulated. - **Part (c):** Using per-unit system confirms load power and source current. --- Let me know if you'd like me to clarify any step!