EDUCRUCIAL

Important questions of JB Gupta book Model Test Paper 2 : (We are working to provide detailed solutions of each question. We are updating this page.) Question 1 : In a magentic material, hysteresis loss takes place primarily due to  (a) rapid reversals of its magnetisation  (b) flux density lagging behind magnetis ing force  (c) molecular friction  (d) its high retentivity Show Answer Ans: D   Question 2 : On which of the following factors does the resolution of a potentiometer depend? (a) Size of wire  (b) Type of contact (c) Composition of wire material  (d) Shape of wire cross-section Show Answer Ans: A   Question 3 : The braking retardation is usually in the range (a) 0.15 to 0.30 km phps  (b) 0.30 to 0.6 km phphs  (c) 0.6 to 2.4 km phps  (d) 3 to 5 km phps  (e) 10 to 15 km phphs Show Answer Ans: D Question 4 : Which of the following methods is used for reduction/elimination of harmonic torques? (a...

  Question : The voltages at V1 and V2 of the arrangement shown in the figure will be respectively A. 6 V and 5.4 V B. 5.4 V and 6 V C. 3 V and 5.4 V D. 6 V and 3 V Solution : For diode D1 it is clear that anode voltage is higher. So it is forward bias. Voltage V2 = 6 - Vt Vt is cut in voltage of diode. V2 = 6 - 0.6 V2 = 5.4 V For diode D2, cathode voltage is higher and hence it is reverse biased.  So V1 = 6 V So option A is correct.  For any further queries please comment below. 

  Question : When a reverse bias is applied to a germanium PN junction diode, the reverse saturation current at room temperature is 0.3m A. Determine the current flowing in the diode when 0.15v forward bias is applied at room temperature. Solution : Given Io = 0.3 * 10-6A and VF = 0.15v The current flowing through the PN diode under forward bias is I = Io (e^(40VF) -1) for germanium diode = 0.3 * 10^(-6) (e^(40*0.15) -1) = 120.73mA. For any further queries please comment below.  

Question : [GSECL JE]  A Uniformly loaded DC distributor is fed at both ends with equal voltages, The voltage drop at the midpoint is...... that of the DC distributor fed at one end. એકસમાન રૂપે ભારિત કરવામાં આવેલા DC વિતરકને બંને છેડેથી સમાન વૉલ્ટેજ પૂરો પાડવામાં આવે છે. મધ્યબિંદુ પરનો વોલ્ટેજ ઘટાડો એક છેડે પૂરા પાડવામાં આવેલા DC વિતરકથી...................છે. A. Twice  B. One-forth C. One - half D. One - third Answer : For distributor fed at one end : Voltage drop for uniformly loaded distributor fed at one end = iR(lx - (x²/2) (Derivation of above equation is given in DC distribution chapter of V. K. Mehta book)  For voltage drop at mid point x = l/2 Substituting x = l/2 Voltage drop at mid point = 3iRl²/8 ...........................(1) For distributor fed at both ends : Voltage drop for uniformly loaded distributor fed at both ends with equal voltages = iR(lx - x²)/2 For voltage drop at mid point x = l/2 Substituting x = l/2 Voltage drop at mid point = iRl²/8 ............

  Question : [GSECL JE ELECTRICAL]  A 1 kVA, single phase, 50 Hz, 100/250 V trarsformer gives the following results on short circuit test conducted at high voltage Side:  Voltmeter reading: 12.5 V, Ammeter reading: 4 A, Wattmeter reading: 80 W.  The equivalent resistance referred to high voltage side will be____.  ઉચ્ચ વોલ્ટેજ બાજુએ હાથ ધરાયેલ લઘુ પરિપથ પરીક્ષણ માટે 1 KVA, એકલચરણ, 50 Hz, 100/250 નું એક પરિવર્તક નીચે મુજબના પરિણામો આપે છે વોલ્ટમીટર વાંચન: 12.5 V, એમીટર વાંચન: 4 A, વૉટમીટર વાંચન: 80 W.  ઉચ્ચ વોલ્ટેજ બાજુએ સંદર્ભિત સમતુલ્ય અવરોધ _______ રહેશે. (a) 20 ohm (b) 5 ohm (c) 40 ohm (d) 10 ohm Answer : (b) 5 ohm To find resistance referred to high voltage side, we need short circuit test data only which is given in question.  High voltage side resistance = Short circuit power/(current)² Resistance referred to HV side = 80 W/ (4 A)² Resistance referred to HV side = 5 ohm.  Option (b) - correct.