Electrical Circuits Objective Questions: Part-3

Electrical Circuits Objective Questions: Part-3

OBJECTIVE QUESTIONS FROM CIRCUIT THEORY:

[1] A balanced RYB-sequence,Y-connected(Star Connected) source with VRN=100 volts is connected to a Δ-connected (Delta connected) balanced load of (8+j6) ohms per phase.Then the phase current and line current values respectively,are[IES2010]
A.10A;30A
B.10√3A;30A
C.10A;10A
D.10√3A;10√3A

 Ans: B

[2] Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is [GATE2012]

        a. 0.8Ω

        b. 1.4Ω

        c. 2Ω

        d. 2.8Ω

Ans:A

[3] The voltage gain A_v of the circuit shown below is [GATE2012]

       a. |A_v| ≈200

       b. |A_v|≈100

       c. |A_v|≈20

       d. |A_v|≈10

Ans:D

[4] If V_A-V_B=6V,  then V_C-V_D is [GATE2012]

       a.  -5V

       b. 2V

       c. 3V

       d. 6V

Ans:A

Solution Hint:
As Va-Vb =6  => I = V/R = 6/2 = 3A
entering current = leaving current [ KCL]
so current from Vd to Vc is same  as 3A
convert the current source to voltage source
V = IR =2 x 1 = 2V

Vd = Vc + 2V + (3x1)V = Vc + 5V
Vc-Vd = -5V

[5]The circuit shown is a [GATE2012]

        a. Low pass filter with f_3db=[1/(R₁+R₂)C] rad/s

        b.High pass filter with f_3db=[1/(R₁C)] rad/s

        c. Low pass filter with f_3db=[1/(R₁C)] rad/s

        d. High pass filter with f_3db=[1/(R₁+R₂)C] rad/s

Ans:B

[6] In the circuit shown, an ideal switch S operated at 100kHz with a duty ratio or 50%. Given thatΔi_c 1.6A peak-to-peak and I₀ is 5A dc, the peak current in S is [GATE2012]

       a. 6.6A

       b. 5.0A

       c. 5.8A

       d. 4.2A

Ans:C 

Solution Hint:
Δi_c = 1.6A peak-to-peak
The positive peak  = Δi_c /2
The current flows through the switch has a peak of =  Io +  Δi_c /2
For your information...The current waveform is like this

The lowest point will be 5A....highest point will be 5+ 0.8

[7] The i-v characteristics of the diode in the circuit given below are 

         i={ [ V- 0.7 /500]A,   V > or = 0.7 V

                                 0A,   V< 0.7 V

The current in the circuit is [GATE 2012]

         a.  10 mA

         b.  9.3 mA

         c.  6.67 mA

         d.  6.2 mA

Ans:D

 

Sol. Hint:
we have to find v and then apply in ct. equ...
KVL...10 = 1K * i + v
10 = 1000i +v = 1000(v-0.7/500) + v
solve it ....v= 0.8...
[8] In the following figure, C₁ and C₂ are ideal capacitors. C₁ has been charged to 12V before the ideal switch S is closed at t=0. The current i(t) for all t is  [GATE 2012]

       a. Zero

       b. A step function

       c. An exponentially decaying function

       d. An impulse function

Ans:D

 

Sol. Hint: 
If it is RC ckt...ie any Resistance in series with capacitor, it will be an exponentially decaying function....here no resistance...so charge instantly....ie impulse function...
[9] The impedance looking into nodes 1 and 2 in the given circuit is [GATE 2012]

   a. 50 Ω

   b. 100 Ω

   c. 5 KΩ

   d. 10.1 KΩ

Ans:A

 

[10] In the circuit given below,the current through the inductor is [GATE 2012]

   

 a.  (2/1+j)A

 b.  (-1/1+j)A

 c.  (1/1+j)A

 d.  0A

Ans:C

 
Sol Hint: consider top half of the circuit
R & L are in parallel with ct. source....other end connected to low potential...ie GND... convert ct. to V source...: