Select Page
Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

## Network Theory MCQ Set 1

1. For the function F (s) = (s2+s+1)/s(s+5)(s+3), after splitting this function into partial fractions, the co-efficient of the term 1/s is?
a) 1/5
b) 1/10
c) 1/15
d) 1/20

Answer: c [Reason:] To obtain the constant A, multiply the given equation with (s) and putting s = 0. The co-efficient of the term 1/s is A= sF (s)|s=0 =(s2+s+1)/(s+5)(s+3) |s=0 =1/15.

2. For the question 1, the co-efficient of 1/(s+5) is?
a) 1.1
b) 2.1
c) 3.1
d) 4.1

Answer: b [Reason:] To obtain the constant B, multiply the given equation with (s+5) and putting s = -5. B= (s + 5)F (s) |s=-5 = (s2+s+1)/s(s+3) |s=-5 =2.1.

3. For the question 1, co-efficient of 1/(s+3) is?
a) -1.17
b) 1.17
c) -2.27
d) 2.27

Answer: a [Reason:] To obtain the constant C, multiply the given equation with (s+3) and putting s = -3. C= (s + 3)F (s)|s= -3 = (s2+s+1)/s(s+5) |s=-3 = -1.17.

4. The partial fraction expansion of the function in question 1 is?
a) 1/15s-2.1/(s+5)+1.17/(s+3)
b) 1/15s-2.1/(s+5)-1.17/(s+3)
c) 1/15s+2.1/(s+5)+1.17/(s+3)
d) 1/15s+2.1/(s+5)-1.17/(s+3)

Answer: d [Reason:] The values of A, B,C are A = 1/15, B = 2.1, C = -1.17. Partial fraction expansion of the function in question 1 is (s2+s+1)/s(s+5)(s+3) =1/15s+2.1/(s+5)-1.17/(s+3).

5. For the function F (s) = (s+5)/s(s2+2s+5) , after splitting this function into the partial fractions, 1/s co-efficient is?
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] To obtain the constant A, multiply the given equation with (s) and putting s = 0. A= sF(s)|s=0 = (s+5)/((s2+2s+5) )=1.

6. For the question 5, the co-efficient of 1/(s+1-j2) is?
a) 1/2
b) -1/2
c) 1/4
d) -1/4

Answer: b [Reason:] To obtain the constant B, multiply the given equation with (s+1-j2) and putting s = -1+j2. B = (s + 1 – j2)F (s)|s= (-1+j2) =(s+5)/s(s+1+j2) |s=-1+j2 = -1/2.

7. For the question 5, determine the co-efficient of 1/(s+1-j2)?
a) -1/4
b) 1/4
c) -1/2
d) 1/2

Answer: c [Reason:] To obtain the constant B*, multiply the given equation with (s+1+j2) and putting s = -1-j2. B* = (s + 1 + j2)F (s)|s= -1 – j2 = (s+5)/s(s+1-j2) |s=-1-j2 = -1/2.

8. The expression of F (s) after splitting into partial fractions in the question 5 is?
a) 1/s-1/2(s+1-j2) -1/2(s+1+j2)
b) 1/s+1/2(s+1-j2) -1/2(s+1+j2)
c) 1/s+1/2(s+1-j2) +1/2(s+1+j2)
d) 1/s-1/2(s+1-j2) +1/2(s+1+j2)

Answer: a [Reason:] The expression of F (s) after splitting into partial fractions in the question 5 is (s+5)/s(s2+2s+5) =1/s-1/2(s+1-j2) -1/2(s+1+j2).

9. The inverse transform of F (s) in the question 5 is?
a) 1+ 1/2 e(-1+j2)t-1/2 e(-1-j2)t
b) 1+ 1/2 e(-1+j2)t+1/2 e(-1-j2)t
c) 1- 1/2 e(-1+j2)t-1/2 e(-1-j2)t
d) 1- 1/2 e(-1+j2)t+1/2 e(-1-j2)t

Answer: c [Reason:] The inverse transform F(s) is f(t), f (t) = L-1(F (s)) = L-1(1/s-1/2(s+1-j2) -1/2(s+1+j2)) =1-1/2 e(-1+j2)t-1/2 e(-1-j2)t.

10. The inverse transform of the function k/(s+a) is?
a) ke-at u(t)
b) keat u(t)
c) ke-at u(t-a)
d) keat u(t-a)

Answer: b [Reason:] The inverse transform of the function k/(s) is k. The inverse transform of the function k/(s+a) is keat u(t). k/(s+a) <—–> keat u(t).

## Network Theory MCQ Set 2

1. Kirchhoff’s Current law is based on law of conservation of
a) energy
b) momentum
c) mass
d) charge

Answer: d [Reason:] Kirchhoff’s current law is based on the law of conservation of charge i.e, charge that flows in = charge that flows out.

2. The current law represents a mathematical statement of fact that
a) voltage cannot accumulate at node
b) charge cannot accumulate at node
c) charge at the node is infinite
d) none of the mentioned

Answer: b [Reason:] Charge cannot accumulate at the node, it can only flow in and out of the node.

3. Kirchhoff’s current law is applied at
a) loops
b) nodes
c) both loop and node
d) none of the mentioned

Answer: b [Reason:] Kirchhoff’s current law can be applicable to nodes only.

4. Determine the current in all resistors in the circuit shown below:

a) 2A, 4A, 11A
b) 5A, 4.8A, 9.6A
c) 9.3A, 20.22A, 11A
d) 10.56A, 24.65A, 14.79A

Answer: d [Reason:] All the resistors are in parallel, so the voltage across each resistor is the same V. i1=V/7 ,i2 =V/3 , i3=V/5.By current law , 50A = V/7 + V/3 + V/5. On solving, we obtain V and then values of i1,i2, i3.

5. For the circuit below , find the voltage across 5Ω resistor and the current through it

a) 1.93 V
b) 2.83 V
c) 3.5 V
d) 5.7 V

Answer: b [Reason:] Here all the resistors are connected in parallel and let the voltage be V. Hence, i15=V/15 , i5=V/5 , i2=V/2 , i1=V/1. By kirchhoff’s current law, V/15 + V/5 + V/2 V/1 +5 = 10. On solving equation, we obtain the value of V. As all resistors are in parallel, voltage across each is same as V.

6. Determine the current through the resistor R3 shown in the figure using KCL

a) 25mA
b) 10mA
c) 20mA
d) 35mA

Answer: a [Reason:] Using KCL , 60mA = 10mA + 25mA + i3.

7. Find the current i3 in the circuit shown below

a) 2A
b) 1A
c) 3A
d) 0.5A

Answer: c [Reason:] By applying the KCL at the node in the circuit , 5A = 2A +i3.

8. Kirchhof’s current law can be mathematically stated as :
a) ∑k=0n I = 0
b) i2k=0n I = 0
c) i∑k=0n I = 0
d) none of the mentioned

Answer: a [Reason:] KCL states that the sum of currents entering and leaving a node is equal to zero.

9. Determine the current if a 20 coulomb charge passes a point in 0.25 seconds
a) 10 A
b) 20 A
c) 2 A
d) 80 A

Answer: d [Reason:] By the definition of electric current, I=q/t.

10. Find the current through the branch containing resistance R3

a) 2A
b) 3.25A
c) 2A
d) 2.75A

Answer: d [Reason:] By KCL, 5A = 0.25A + 2A + i3.

## Network Theory MCQ Set 3

1. Kirchhoff’s voltage law is based on principle of conservation of
a) energy
b) momentum
c) mass
d) charge

Answer: a [Reason:] KVL is based on the law of conservation of energy.

2. In a circuit with more number of loops, which law can be best suited for the analysis?
a) KCL
b) Ohm’s law
c) KVL
d) None of the mentioned

Answer: c [Reason:] KVL can be best suited for circuits with more number of loops.

3. Determine the unknown voltage drop in the circuit below
a) 11V
b) 10V
c) 19V
d) 5V

Answer: c [Reason:] By applying KVL to the loop, we get, 2 + 1 + V + 3 + 5 – 30 = 0.

4. Determine V in the circuit
a) 28.8V
b) 34.4v
c) -28.8V
d) 28V

Answer: c [Reason:] First, apply KCL to the second node i.e, i6 + I = i8 . Next, apply KVL to the first loop i.e, -20 + V2 + V5 =0. But V2 = 7*2=14V. Obtain V5 and I5=V5/5. Again apply KVL to central loop i.e, -V5 + V6 + V = 0. Apply KCL to the first node again, seek the value of V6 and finally we get V.

5. Find V and I in the circuit
a) -39V , -4.875A
b) 39V , -4.875A
c) -39v , 4.875a
d) 39V , 4.875A

Answer: a [Reason:] By applying KVL to the loop, we get, +30+9-V = 0. then, I=V/8.

6. Mathematically, Kirchhoff’s Voltage law can be as
a) ∑_(k=0)n(V) = 0
b) V2∑_(k=0)n(V) = 0
c) V∑_(k=0)n(V) = 0
d) none of the mentioned

Answer: a [Reason:] According to KVL, the sum of all voltages of branches in a closed loop is zero.

7. Determine the value of V and the power supplied by the independent current source
a) 20V , 300mw
b) 27V , 498mW
c) 26.6v , 532mW
d) 25V , 322mW

Answer: c [Reason:] Apply KCL to the node, we get i – 2i1 – 0.02 – i1 = 0. Next, apply ohm’s law to each resistor, i=v/4000 and i1= -v/6000 and substitute in the above equation. For the power supplied by the independent source P = V*0.02A.

8. Determine V in the circuit
a) -11.6V
b) 23.2V
c) -23.2V
d) 11.6V

Answer: a [Reason:] First, apply KCL to the second node i.e, i8 + I = i10 . Next, apply KVL to the first loop i.e, -20 + V i1 + V6 =0. But V4 = 4*4=16V. Obtain V6 and I6=V6/6. Again apply KVL to central loop i.e, -V6 + V8 + V = 0. Apply KCL to the first node again, seek the value of V8 and finally we get V.

9. Find V and I in the circuit
a) 19V, 0.0633A
b) -19V, 0.0633A
c) 19V, -0.0633A
d) -19V,- 0.0633A

Answer: a [Reason:] By applying KVL to the loop, we get, -12-7+V = 0. then, I=V/300.

## Network Theory MCQ Set 4

1. The current in a closed path in a loop is called?
a) loop current
b) branch current
d) twig current

Answer: c [Reason:] In a loop there exists a closed path and a circulating current which is called link current. The current in any branch can be found by using link currents.

2. Tie-set is also called?
a) f loop
b) g loop
c) d loop
d) e loop

Answer: a [Reason:] The fundamental loop formed by one link has a unique path in the tree joining the two nodes of the link. This loop is also called f-loop.

3. Consider the graph shown below. If a tree of the graph has branches 4, 5, 6, then one of the twigs will be?

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

Answer: d [Reason:] Branches of the tree are called twigs. So 4, 5, 6 are the twigs of the tree. The current in any branch of a graph can be found by using link currents.

4. Consider the graph shown in the question 3 above. If a tree of the graph has branches 4, 5, 6, then one of the links will be?
a) 3
b) 4
c) 5
d) 6

Answer: a [Reason:] The branches of the co-tree are called links. So the links will be 1, 2, 3. For a given tree of a graph addition of each link between any two nodes form a loop called fundamental loop.

5. The loop current direction of the basic loop formed from the tree of the graph is?
a) same as the direction of the branch current
b) opposite to the direction of the link current
c) same as the direction of the link current
d) opposite to the direction of the branch current

Answer: c [Reason:] The loop current direction of the basic loop formed from the tree of the graph is same as the direction of the link current.

6. Consider the graph shown below. The direction of the loop currents will be? (ACW – Anticlockwise, CW – Clockwise).

a) I1 ACW
b) I2 ACW
c) I3 CW
d) I4 ACW

Answer: a [Reason:] The direction of the loop current will be along the direction of the link current in a basic loop. So I1 – ACW, I2 – CW, I3 – ACW, I4 – CW.

7. For Tie-set matrix, if the direction of current is same as loop current, then we place ___ in the matrix.
a) +1
b) -1
c) 0
d) +1 or -1

Answer: a [Reason:] For Tie-set matrix, if the direction of current is same as loop current, then we place +1 in the matrix.

8. If a row of the tie set matrix is as given below, then its corresponding equation will be?
1 2 3 4 5 6 7 8
I1 -1 +1 0 0 +1 0 0 0
a) -V1+V2+V3=0
b) -I1+I2+I3=0
c) -V1+V2-V3=0
d) -I1+I2-I3=0

Answer: a [Reason:] KVL equations are derived from tie set matrix and these include voltages not currents. So, -V1+V2+V3=0.

9. The matrix formed by link branches of a tie set matrix is?
a) Row matrix
b) Column matrix
c) Diagonal matrix
d) Identity matrix

Answer: d [Reason:] As the direction of the basic loops of the tree are taken along the direction of the link currents, then the matrix formed by the link currents will be a identity matrix.

10. The number of tie set matrices formed from a graph are?
a) NN-1
b) NN
c) NN-2
d) NN+1

Answer: c [Reason:] For every tree, there will be a unique tie set matrix. So there will be NN-2 tie set matrices.

## Network Theory MCQ Set 5

1. The maximum power is delivered from a source to its load when the load resistance is ______ the source resistance.
a) greater than
b) less than
c) equal to
d) less than or equal to

Answer: c [Reason:] The maximum power is delivered from a source to its load when the load resistance is equal to the source resistance. The maximum power transfer theorem can be applied to both dc and ac circuits.

2. If source impedance is complex, then maximum power transfer occurs when the load impedance is _______ the source impedance.
a) equal to
b) negative of
c) complex conjugate of
d) negative of complex conjugate of

Answer: c [Reason:] The maximum power transfer theorem can be applied to complex impedance circuits. If source impedance is complex, the maximum power transfer occurs when the load impedance is complex conjugate of the source impedance.

3. If the source impedance is complex, then the condition for maximum power transfer is?
a) ZL = ZS
b) ZL = ZS*
c) ZL = -ZS
d) ZL = -ZS*

Answer: b [Reason:] The maximum power is transferred when the load resistance is equal to the source resistance. The condition for maximum power transfer is ZL = ZS*.

4. If ZL = ZS*, then?
a) RL = 1
b) RL = 0
c) RL = -RS
d) RL = RS

Answer: d [Reason:] If ZL = ZS*, then RL = RS. This means that the maximum power transfer occurs when the load impedance is equal to the complex conjugate of source impedance ZS.

5. For ZL = ZS*, the relation between XL and XS is?
a) XL = XS
b) XL = 0
c) XL = 1
d) XL = -XS

Answer: d [Reason:] For ZL = ZS*, the relation between XL and XS is XL = -XS. Maximum power transfer is not always desirable since the transfer occurs at a 50 percent efficiency.

6. In the circuit shown below, find the value of load impedance for which source delivers maximum power.

a) 15-j20
b) 15+j20
c) 20-j15
d) 20+j15

Answer: a [Reason:] The maximum power transfer occurs when the load impedance is equal to the complex conjugate of source impedance ZS. ZL = ZS* = (15-j20) Ω.

7. The load current in the circuit shown in the question 6 is?
a) 1.66∠90⁰
b) 1.66∠0⁰
c) 2.66∠0⁰
d) 2.66∠90⁰

Answer: b [Reason:] The load current is the ratio of voltage to the impedance. So the load current is I=(50∠0o)/(15+j20+15-j20) =1.66∠0o A.

8. The maximum power delivered by the source in the circuit shown in the question 6 is?
a) 39.33
b) 40.33
c) 41.33
d) 42.33

Answer: c [Reason:] The term power is defined as the product of the square of current and the impedance. So the maximum power delivered by the source in the circuit is P = I2RxZ = 1.662×15 = 41.33W.

9. For the circuit shown, the resistance R is variable from 2Ω to 50Ω. What value of RS results in maximum power transfer across terminals ‘ab’.

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