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T5 1 ph-ac

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T5 1 ph-ac

1. 1. ELE101/102 Dept of E&E,MIT Manipal 1 Tutorial 1. A resistance of 50Ω is connected in series with an inductance of 100 mH across a 230V, 50 Hz, single phase AC supply. Calculate a) Impedance b) current drawn c) power factor d) power consumed e) Draw the phasor diagram. Ans: Ω°∠ 14.3259 A14.32898.3 °−∠ 0.847 lag 759.15W IVR VL V 32.12º
2. 2. ELE101/102 Dept of E&E,MIT Manipal 2 Tutorial 2. A resistance of 50Ω is connected in series with a capacitance of 100 µF across a 230V, 50 Hz, single phase AC supply. Calculate a) Impedance b) current drawn c) power factor d) power consumed e) Draw the phasor diagram. Ans: Ω°−∠ 48.32272.59 A°∠ 48.3288.3 0.843 lead 752.81 W I VR VC V 32.48º
3. 3. ELE101/102 Dept of E&E,MIT Manipal 3 Tutorial 3. The value of the capacitor in the circuit given below is 20 µF and the current flowing through the circuit is 0.345 A. If the voltages are as indicated, find the applied voltage, the frequency and loss in the coil. C R L RLI V 25 V 40 V 55 V coil 50 V degree26.768byVleadsIcurrent34.155V;V50Hz,f :Ans == Power Loss = 1.8967 W
4. 4. ELE101/102 Dept of E&E,MIT Manipal 4 Tutorial 4. An emf of v= 326 sin 418t is applied to a circuit. The current is i = 20 sin(418t + 60). Find the circuit components, frequency of the input voltage and power factor. Solution: f=66.5 Hz pf = 0.5. Z = Vm / Im = 16.3 Ω. R = Z cos φ; R = 8.15 Ω XC = 14.11 Ω; C= 169.6 microfarad
5. 5. ELE101/102 Dept of E&E,MIT Manipal 5 Tutorial 5. A current of 5 A flows through a non inductive resistance in series with a coil when supplied at 250 V, 50 Hz. If the voltage across the resistance is 125 V and that across the coil is 200V, calculate a) the impedance, reactance and resistance of the coil. b) power absorbed by the coil. c) Total power. Draw the phasor diagram. R RL L L coil 125 V 200 V 250 V, 50 Hz
6. 6. ELE101/102 Dept of E&E,MIT Manipal 6 Tutorial Phasor Diagram IVR = IR IRL IXL V Vcoil R = 25 Ω Zcoil = 40 Ω RL = 5.5 Ω XL = 39.62 Ω Total pf = 0.61 Total power = 762.5 W Power absorbed by the coil = 137.5 W
7. 7. ELE101/102 Dept of E&E,MIT Manipal 7 Tutorial 6. Find the values of R and C so that Vb = 3 Va and Vb and Va are in quadrature. Find also the phase relation between V and Va , Va and I. 0.0255 H 6 Ω R C I Vb Va V=240V, 50 Hz I - ref Vb Va θa θb=53.16º Zb = 10∠53.16º Since Vb and Va are in quadrature. Za = 3.336∠-90+53.16º = 3.336∠-36.84º R = 2.669 Ω FC 3 1059.1 − ×= Za = Zb / 3 = 3.336 Solution