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Tutorial

Tutorial. 1) Find the Average value and RMS value of the given non sinusoidal waveform shown below. A. 0. T. T/4. t. T/2. -A. Ans: Iav= A / 2. V. V L. 32.12 º. I. V R. Tutorial.

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Tutorial

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  1. Tutorial 1) Find the Average value and RMS value of the given non sinusoidal waveform shown below. A 0 T T/4 t T/2 -A Ans: Iav= A / 2 Dept of E&E MIT Manipal

  2. V VL 32.12º I VR Tutorial 2. A resistance of 50 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: 0.847 lag 759.15W Dept of E&E,MIT Manipal

  3. VR I 32.48º VC V Tutorial 3. A resistance of 50 is connected in series with a capacitance of 100 F 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: 0.843 lead 752.81 W Dept of E&E,MIT Manipal

  4. coil C R L RL I 25 V 55 V V 40 V 50 V Tutorial 4. The value of the capacitor in the circuit given below is 20 F 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. Power Loss = 1.8967 W Dept of E&E,MIT Manipal

  5. Tutorial • 5. 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 . • R = Z cos ; R = 8.15  • XC = 14.11 ; C= 169.6 microfarad Dept of E&E,MIT Manipal

  6. coil L RL R L 200 V 125 V 250 V, 50 Hz Tutorial • 6. 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, calculate a) the impedance, reactance and resistance of the coil. b) power absorbed by the coil. c) Total power. Draw the phasor diagram. Dept of E&E,MIT Manipal

  7. coil L RL R L V IXL Vcoil VR = IR IRL I 200 V 125 V 250 V, 50 Hz Tutorial Zcoil = 40  R = 25  XL = 39.62  RL = 5.5  Total pf = = 0.61 Phasor Diagram Total power = 762.5 W Power absorbed by the coil = 137.5 W Dept of E&E,MIT Manipal

  8. Za = Zb / 3 = 3.336 6  C R 0.0255 H I Va Vb Vb V=240V, 50 Hz b=53.16º I - ref a Va 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. Solution Zb = 1053.16º Since Vb and Va are in quadrature. Za = 3.336-90+53.16º = 3.336-36.84º R = 2.669  Dept of E&E,MIT Manipal

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