1. ELE 102/102Dept of E&E MIT Manipal1 Tutorial 1) Find the Average value and RMS value of the given non sinusoidal waveform shown below. 0 A -A T/4 T/2 T t Ans: Iav= A / 2

2. ELE101/102Dept of E&E,MIT Manipal2 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 I VRVR VLVL V 32.12º

3. ELE101/102Dept of E&E,MIT Manipal3 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 I VRVR VCVC V 32.48º

4. ELE101/102Dept of E&E,MIT Manipal4 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. C RL RLRL I V 25 V 40 V 55 V coil 50 V Power Loss = 1.8967 W

5. ELE101/102Dept of E&E,MIT Manipal5 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 = V m / I m = 16.3 . R = Z cos  ; R = 8.15  X C = 14.11  ; C= 169.6 microfarad

6. ELE101/102Dept of E&E,MIT Manipal6 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. R RLRL L L coil 125 V 200 V 250 V, 50 Hz

7. ELE101/102Dept of E&E,MIT Manipal7 Tutorial Phasor Diagram R RLRL L L coil 125 V 200 V 250 V, 50 Hz I V R = IR IR L IX L V V coil R = 25  Z coil = 40  R L = 5.5  X L = 39.62  Total pf == 0.61 Total power = 762.5 W Power absorbed by the coil = 137.5 W

8. ELE101/102Dept of E&E,MIT Manipal8 Tutorial 6. Find the values of R and C so that V b = 3 V a and V b and V a are in quadrature. Find also the phase relation between V and V a, V a and I. 0.0255 H 6  R C I VbVb VaVa V=240V, 50 Hz I - ref VbVb VaVa aa  b =53.16º Z b = 10  53.16º Since V b and V a are in quadrature. Z a = 3.336  -90+53.16º = 3.336  -36.84º R = 2.669  Z a = Z b / 3 = 3.336 Solution