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RRB JE Stage II : Electrical Engineering Quiz | 17th August

Dear aspirants,

RRB JE Stage II exam is scheduled to be conducted from 28th August to 1st September2019. The overall merit for the selection will be made on the basis of the marks scored in stage II, so candidates need to prepare thoroughly for the second stage. To help the aspirants, Adda247 has come up with a study plan in which we will provide you daily quizzes of all technical subjects. The below quiz will be helpful for the candidates who opted Electrical as the exam group in RRB JE stage II. 

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Q1. If three 15 μ F capacitors are connected in series, the net capacitance is


(a) 5 μF
(b) 30 μF
(c) 45 μF
(d) 50 μF

Q2. The current ‘I’ in the electric circuit shown below is ?

(a) 3.7 A
(b) 1A
(c) 2.7 A
(d) 1.7 A

Q3. “The total electric flux through any closed surface surrounding charges is equal to the amount of charge enclosed”.The above statement is associated with
(a) Coulomb’s square law
(b) Gauss’s law
(c) Maxwell’s first law
(d) Maxwell’s second law

Q3. “The total electric flux through any closed surface surrounding charges is equal to the amount of charge enclosed”.
The above statement is associated with
(a) Coulomb’s square law
(b) Gauss’s law
(c) Maxwell’s first law
(d) Maxwell’s second law

Q4. Total capacitance between the points L and M in figure is :

Q5. The sparking between two electrical contacts can be reduced by inserting a
(a) capacitor in parallel with contacts
(b) capacitor in series with each contact
(c) resistance in line
(d) none of the above

S5. (a)
An arc between two electrodes can be initiated by ionization of atmospheric air. If we connect a capacitor across the contacts of the electrodes, there is a chance to make a short circuit by capacitor. Hence, there is no air gap between contacts and thereby sparking can be reduced.

Q6. The value of V in the circuit shown in the given figure is

(a) 1V
(b) 2V
(c) 3V
(d) 4V

Q8. In the figure shown below, if we connect a source of 2-V, with internal resistance of 1Ω at AA’ with positive terminal at A, then current through R is

(a) 2A
(b) 1.66 A
(c) 1A
(d) 0.625 A

Q10.Time constant of the network shown in figure is

 

 

 

 

 

 

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