##
__Objective Questions from Circuit Theory:__

[1] A balanced RYB-sequence,Y-connected(Star Connected) source with V_{RN}=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

(e) None of the above

**Answer: 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

(e) None of the above

[3] The voltage gain A

**Answer: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
(e) None of the above

**Answer:D**

[4] If V

_{A}-V

_{B}=6V, then V

_{C}-V

_{D}is [GATE2012]

(a) -5V

(b) 2V

(c) 3V

(d) 6V

(e) None of the above

**Answer:A**

[5]The circuit shown is a [GATE2012]

(a) Low pass filter with f

_{3db}=[1/(R_{1}+R_{2})C] rad/s
(b) High pass filter with f

_{3db}=[1/(R_{1}C)] rad/s
(c) Low pass filter with f

_{3db}=[1/(R_{1}C)] rad/s
(d) High pass filter with f

_{3db}=[1/(R_{1}+R_{2})C] rad/s
(e) None of the above

**Answer: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 I

_{0}is 5A dc, the peak current in S is [GATE2012]

(a) 6.6A

(b) 5.0A

(c) 5.8A

(d) 4.2A

(e) None of the above

**Answer: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

(b) 9.3 mA

(c) 6.67 mA

(d) 6.2 mA

(e) None of the above

**Answer: D**

**Solution 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, C

(a) ZeroKVL...10 = 1K * i + v

10 = 1000i +v = 1000(v-0.7/500) + v

solve it ....v= 0.8...

[8] In the following figure, C

_{1}and C_{2}are ideal capacitors. C_{1}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]
(b) A step function

(c) An exponentially decaying function

(d) An impulse function

(e) None of the above

**Answer: D**

**Solution 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

(e) None of the above

**Answer: 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

(e) None of the above

**Answer: A**

**Solution 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...

Sir,this blog is so much helpful..and one small request sir,

ReplyDeleteplease provide solutions also for the questions sir.

Thank u

Thanks for your comments...ans hint is added...For future post we will give solutions hint...

DeleteFor previously published posts, please let us know the question no, we will add the soultion hint for that answer.

Deletethanks sir for your valuable efforts .......

ReplyDeleteIt may be great if you add next post link at end of every post .......

Make playlist like youtube

thanks again

sir please provide objective type for power quality engineering also and also give explanations

ReplyDeletereally it is very useful to all students thank u sir

ReplyDeleteAnswer is C for Question No. 10

ReplyDeletethe question no 10 answer is C

ReplyDelete