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HOME WORK # 3 – SOLUTIONS 1. An ideal transformer has an input voltage of 480 V. The output current and voltage are 10 A and 120 V. Determine the value of input current. Solution: I1 = V2 I 2 (120 )(10 ) = = 2.5 A V1 480 2. A 15-kVA, 2400:240-V, 60 Hz transformer has the following equivalent circuit parameters: R1 = 2.5Ω Rc = 32k Ω R2 = 0.025Ω X 1 = 7.0Ω X m = 11.5k Ω X 2 = 0.070Ω If the transformer is supplying a 10-kW, 0.8 PF lagging load at rated voltage, calculate: (a) (b) (c) Input voltage Input current, and Input power factor Solution: Reflect R2 and X 2 to the primary side:. 2 ⎛N ⎞ ⎛ 2400 ⎞ R2′ = ⎜ 1 ⎟ R2 = ⎜ ⎟ ( 0.025 ) = 2.5 Ω ⎝ 240 ⎠ ⎝ N2 ⎠ 2 2 ⎛N ⎞ ⎛ 2400 ⎞ X 2′ = ⎜ 1 ⎟ X 2 = ⎜ ⎟ ( 0.070 ) = 7.0 Ω N ⎝ 240 ⎠ ⎝ 2⎠ 2 (a,b) Assume V2 on reference; then, aV2 = (10 )( 240∠0° ) = 2400∠0° V The load current is found by I2 = P2 10,000 = = 52.08 A V2 ( PF ) 240 ( 0.8 ) I 2 = I 2 ∠ − cos −1 ( 0.8 ) = 52.08∠ − 36.87° A The reflected load current is I 2′ = 1 I 2 = 5.208∠ − 36.87° A 10 The excitation branch voltage is found by use of KVL. E = I 2′ ( R2′ + jX 2′ ) + aV2 = ( 5.208∠ − 36.87° )( 7.433∠70.35° ) + 2400∠0° E = 2432.381∠0.50° V The excitation branch current follows as Io = Ic + Im = E E 2432.381∠0.50° 2432.381∠0.50° + = + Rc jX m 32,000 11,500∠90° I o = 0.224∠ − 69.70° A The input current can now be determined by KCL. I1 = I 2′ + I o = 5.208∠ − 36.87° + 0.224∠ − 69.70° I1 = 5.398∠ − 38.16° A Application of KVL yields the input voltage. V1 = I1 ( R1 + jX 1 ) + E = 5.398∠ − 38.16° ( 7.433∠70.35° ) + 2432.381∠0.50° V1 = 2466.61∠0.99° V V1 = 2466.61 V (c) With V1 and I1 known, the input power factor is given by PFin = cos ( ∠V1 − ∠I1 ) = cos ( 0.99° + 38.16° ) = 0.775 lagging 3. A 20 kVA, 2500/250 V, 50Hz, single-phase transformer gave the following test result: Open-circuit test: 250V, 1.5A, and 100 watts measured on the low voltage side Short-circuit test: 100V, 8A, and 320 watts measured on the on high voltage side a) Compute the parameters of the equivalent circuits and draw the equivalent circuit of the transformer showing all the values. b) Calculate the equivalent circuit of the transformer in per unit. Solution: From the given test results data, we have P Rwdg = sc2 = 5Ω I sc 2 ⎛V ⎞ X l = ⎜ sc ⎟ − Rwdg 2 = 12Ω ⎝ I sc ⎠ and Rcore (V ) = oc Poc 2 = 625Ω 2 2 ⎛ V ⎞ ⎛V ⎞ I oc = ⎜ oc ⎟ + ⎜ oc ⎟ ⎝ Rcore ⎠ ⎝ X m ⎠ Therefore, X m = 173Ω . Rcore and Xm are low-voltage side values, which can be converted to the high voltage side: R_core = Rcore *(N1/N2)2 = 62.5 kΩ, and X_m = Xm *(N1/N2)2 = 17.3 kΩ. We can find the inductances: Xl Ll = = 38.2mH 100π X Lm = m = 550.68mH 100π (b) Equivalent circuit in per unit: 25002 = 312.5Ω 20k 2502 Z base 2 = = 3.125Ω 20k 625 Rcore , pu = = 200, pu Z base 2 173 X m , pu = = 55.36, pu Z base 2 12 X l , pu = = 0.0384, pu Z base1 5 Rwdg , pu = = 0.016, pu Z base1 Z base1 =