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46) 46) In a Y-connected source feeding a A-connected load: A) each phase of the load has one-third of the full line voltage across it. b) each phase of the load has a voltage across it equal to V times the line voltage C) each phase of the load has the full line voltage across it. D) each phase of the load has two-thirds of the full line voltage across it. 47) 47) A three-phase generator is connected to three 702 loads. Each coil generates 120 V ac. A common neutral line exists. How much current is flowing through the common neutral line? A) OA B) 2.26 A C) 5.13 A D) 3.42 A 48) 48) In a A-connected source feeding a A-connected load, the: A) load voltage and line voltage cancel for a given phase. B) load voltage and line voltage are all equal for a given phase. C) load voltage and line voltage are one-third the source voltage for a given phase. D) load voltage and line voltage are two-thirds the source voltage for a given phase. 49) In a certain A-connected balanced load, with line voltages of 230 V and the impedances are 60 235° 12, determine the magnitude of the load currents. A) 2.82 A B) 7.19 A C) 6.64 A D) 4.82 A 50) 30) The power in each phase of a balanced three-phase system is 1500 W. What is the total power? A) 1500 W B) 4500 W C) 500 W D) 2278 W

Answer 1

în a y connected sowice feeling a s. conneded load A) each phase of the load has one third of the tall line voltage across it. B) each phase of the load has a voltage across it ealed to 3 times of the line voltage e) each phase of the load has the full line voltage a cyoss it. d) each phase of the load has two third of the full line voltage across it Solution: ti LOAD Source 4 V-128 NW - VRY Star connected 3-4 Source Delta Connected Load. It we observe the above diagrame VR Leo, Vy L120, Volzug supply voltages in each phase So Live voltage VR-Vy = vey This vry is applied across phase of a load impedance (2) between the points Ry so The phase impedance "z" of a load having line cs Scanned vequivalerat voltage of VRY. CamScanner

Also the load is in delta Connection so in della the line voltage = Phose voltage so Answell is option (C): Each phase of the load hos The full line voltage across it. Q (47): A Three Phase generator is connected to Three your loads: Each Coil generates 120 V ac. A common neutral line exists. How much carrent is flowing Through The common neutral line ? VADOA B) 2.26€ )5:13A D3:42A Solution. A common neatral exists between source and load so source and load must be star Connected of shown below 1200 120V 60° L-2409 "Neutral line" - Source y 120V L-1200 O' Load 1 Ton * The source and load's both are balanced. So The current in the neutral wire cor) live is zero why because the potential CS Scanned at ito , o' is same. CamScanner So AnswJ7 = (A) OA. DA

&(48): In a - connected source feeding a - A connected load The A load voltage and line voltage cancel for a given phase VB) load voltage and line voltage are equal for a given phase () load voltage and line voltage are one-third The Source voltage for a given phase. D) load voltage and line voltage are Two-third The source voltage for a given phase. Solution TE COMO Source vs @ Olle -VRY Delta connected Delta connected Source COA) If we obsolue above diagram, The load for phase voltage = Live voltage (detta connection) Also Vey load voltage = VRY Line voltage so absolutely load voltage and line voltage are equal for a given phase Ury so The Answell option (B): load voltage and Scanned with CamScannerline voltage are equal for a given Phaye.

& U): In a certain D-connected balanced load with line voltage of 230 V and impedance are 60 L35 r determine the magnitude of the load currents. A) 2.82 A B) 7.19 A B 6.644 19482 A "A-Connected LOAD" Solution! 230 V Line hier here you colerefter a colorir voltage 2 60 (351 * load is balanced so each phase has equal jumpedance. Also each pharge of load Carry equal amount of current also so here we consider Ry phase. Ry phase current I phase = Uphage but ty detox In delta connection V line = Uphase so Ipnose - Vline : 230 z oo (350 - 5 8 35 A (35°. Here we want Load current nothing but Live current * in delta connection I live = V3 I phase Scanned with CamScanner

I line cuircut : v3 ( 3.833 ) 36.6389 I line = 6.64 A = I load so The Answer: option: (C) 6.64 A Q :(50): The power in each phase of a balanced Three-phase system is 15oow. What is the Total Power A) 1500 W ws) 4500 W c) 500W D) 2278 w Solutions c. In a Balanced load îf phase voltage is :v . Phase current is - I Power factor of phase is = cosa per phase Power is = vI cos & K phase Three Phase fowar is = 3v1 Coso k 3-0 1500 W for so here each phase Power is a balanced load. So VI Cosp = 1500 W Total Power (3ui cosd) = 1500 x 3 Total 3-8 lower = 4500 W The Answer: option (B): 4500 W so Scanned with CamScanner