Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

Q31: The current through the 2 kΩ resistance in the circuit shown is      (2009)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 0 mA
(b) 1 mA
(c) 2 mA
(d) 6 mA
Ans:
(a)
Sol: Bridge is balanced i.i. node C and node D are at same potential. Therefore, no current flows through 2kΩ resistor.

Q32: In the circuit shown in the figure, the value of the current i will be given by     (2008)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 0.31 A
(b) 1.25 A
(c) 1.75 A
(d) 2.5 A
Ans:
(b)
Sol: Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)By KVL in loop-1,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)By KVL in loop-2,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q33: Assuming ideal elements in the circuit shown below, the voltage Vab will be     (2008)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) -3 V
(b) 0 V
(c) 3 V
(d) 5 V
Ans:
(a)
Sol: i = 1A
Applying KVL,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)NOTE: KVL is based on the conservation of energy.

Q34: A 3 V DC supply with an internal resistance of 2 Ω supplies a passive non-linear resistance characterized by the relation Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE). The power dissipated in the non linear resistance is        (2007)
(a) 1.0 W
(b) 1.5 W
(c) 2.5 W
(d) 3.0 W
Ans:
(a)
Sol: Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)-3 A is rejected, because the non-linear resistor is passive and the only active element in the circuit is 3 V DC supply. Which is supplying the power to the resistor.
So, INL= 1A
Power dissipated in the non-linear resistor 
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q35: In the figure given below the value of R is     (2005)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 2.5 Ω
(b) 5.0 Ω
(c) 7.5 Ω
(d) 10.0 Ω
Ans:
(c)
Sol: The resultant (R) when viewed from voltage source = 100/8  = 12.5
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q36: The rms value of the resultant current in a wire which carries a dc current of 10 A and a sinusoidal alternating current of peak value 20 A is      (2004)
(a) 14.1 A
(b) 17.3 A
(c) 22.4 A
(d) 30.0 A
Ans:
(b)
Sol: R.M.S value of d.c current = 10A = Idc
R.M.S. value of sinusoidal current Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
R.M.S. value of resultant,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q37: In figure, the value of resistance R in Ω is    (2004)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 10
(b) 20
(c) 30
(d) 40
Ans:
(b)
Sol: Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q38: In figure, Ra, Rb and Rc are 20Ω, 10Ω and 10Ω respectively. The resistances R1, R2 and R3 in Ω of an equivalent star-connection are        (2004)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 2.5, 5, 5
(b) 5, 2.5, 5
(c) 5, 5, 2.5
(d) 2.5, 5, 2.5
Ans: 
(a)
Sol:
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Remember: If all the branches of Δ-connection has same impedance Z, then impedance of branch of Y-connection be Z/3.

Q39: In figure, the value of the source voltage is      (2004)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 12 V
(b) 24 V
(c) 30 V
(d) 44 V
Ans:
(c)
Sol: Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)

Method-1:
Using KCL,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Solving equation (i) and (ii), we get
Va = 18V and E = 30V
Method-2:
I = 2+1 = 3A
Applying KVL in second loop,
E = 2 x 6 + 3 x 6 = 30V

Q40: In figure, the value of R is    (2003)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 10 Ω
(b) 18 Ω
(c) 24 Ω
(d) 12 Ω
Ans: 
(d)
Sol: By KCL,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Potential difference between node x and y = 60V
By taking KCL at node y
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q41: In figure, the potential difference between points P and Q is     (2003)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 12 V
(b) 10 V
(c) -6 V
(d) 8 V
Ans: 
(c)
Sol: Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)From equation (i),
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)From equation (ii),
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q42: A segment of a circuit is shown in figure. VR = 5V,  VC = 4sin2t. The voltage VL is given by    (2003)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 3 - 8 cos 2t
(b) 32 sin 2t
(c) 16 sin 2t
(d) 16 cos 2t
Ans:
(b)
Sol: By KCL,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)NOTE: KCL is based on the law of conservation of charges.

Q43: Figure Shows the waveform of the current passing through an inductor of resistance 1Ω and inductance 2 H. The energy absorbed by the inductor in the first four seconds is     (2003)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 144 j
(b) 98 j
(c) 132 j
(d) 168 j
Ans:
(c)
Sol: For 0 < t < 2s current varies linearly with time and given as i(t) = 3t and for 2s < t < 4s current is constant, i(t) = 6A.
The energy absorbed by the inductor (Resistance neglected) in the first 2 sec,
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)The energy absorbed by the inductor in (2 → 4) second
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)A pure indictor does not dissipate enegy but only stores it. Due to resistance, some energy is dissipated in the resistor. Therefore, total energy stored in the inductor and the energy dissipated in the resistor.
The energy dissipated by the resistance in 4 sec.
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)The total energy absorbed by the inductor in 4 sec
= 96J + 36J = 132J

Q44: Consider the star network shown in figure. The resistance between terminals A and B with terminal C open is 6Ω, between terminals B and C with terminal A open is 11Ω, and between terminals C and A with terminal B open is 9Ω. Then     (2001)
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 𝑅𝐴RA = 4Ω, RB = 2Ω, R= 5Ω
(b) 𝑅𝐴RA = 2Ω, RB = 4Ω, R= 7Ω
(c) 𝑅𝐴RA = 3Ω,  RB = 3Ω,  R= 4Ω
(d) 𝑅𝐴RA = 5Ω, RB = 1Ω,  RC = 10Ω
Ans:
(b)
Sol: When C is open, Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
When B is open, Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
When A is open,Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
On solving the above equations
Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q45: Two incandescent light bulbs of 40 W and 60 W rating are connected in series across the mains. Then       (2001)
(a) the bulbs together consume 100 W
(b) the bulbs together consume 50W
(c) the 60 W bulb glows brighter
(d) the 40 W bulb glows brighter
Ans:
(d)
Sol: Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Therefore , resistance of 40 W bulb > resistance of 60 W bulb.
For series connection, current through both the bulbs will be same P = I2R (for series connection).
Power consumed by 40 W bulb > Power consumed by 60 W bulb.
Hence the 40 W bulb glows brighter. 

The document Previous Year Questions- Basics - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Network Theory (Electric Circuits).
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