Questions & Answers On Useful Theorems in Circuit Analysis
Questions & Answers On Useful Theorems in Circuit Analysis
Questions & Answers On Useful Theorems in Circuit Analysis
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Answer: a
Explanation: After transformation to star, the resistor at
node X is RxRy/( Rx+Ry+Rz) and this resistance lies
between Rx, Ry in star connection.
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c) RxRy/(Rx+Ry+Rz)
d) (Rz+Ry)/(Rx+Ry+Rz)
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Answer: b
Explanation: After transformation to star, the resistor at
node Y is RzRx/(Rx+Ry+Rz) and this resistance lies
between Rx, Rz in star connection.
Answer: c
Explanation: After transformation to star, the resistor at
node Y is RzRy/(Rx+Ry+Rz) and this resistance lies
between Rz, Ry in star connection.
Answer: a
Explanation: After transformation to delta, the resistance
between 1 and 2 in delta connected system will be (R1R2+
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Answer: c
Explanation: After transformation to delta, the resistance
between 2 and 3 in delta connected system will be (R1R2+
R2R3+ R3R1)/R1 and this resistance lies between R3, R2 in
delta connection.
Answer: c
Explanation: After transformation to delta, the resistance
between 2 and 3 in delta connected system will be (R1R2+
R2R3+ R3R1)/R2 and this resistance lies between R1, R3 in
delta connection.
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a) 1
b) 2
c) 3
d) 4
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Answer: d
Explanation: Performing delta to star transformation we
obtain the equivalent resistance at node A is
R=(11×12)/(11+12+13)=4Ω.
Answer: b
Explanation: Performing delta to star transformation we
obtain the equivalent resistance at node A is
R=(12×13)/(11+12+13)=4.66Ω.
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a) 30
b) 31
c) 32
d) 33
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Answer: c
Explanation: The equivalent resistance between node 1 and
node 3 in the star connected circuit is
R=(10×10+10×11+11×10)/10=32Ω.
Answer: b
Explanation: The equivalent resistance between node 1 and
node 3 in the star connected circuit is
R=(10×10+10×11+11×10)/11=29Ω.
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Answer: b
Explanation: In Superposition theorem, while considering a
source, all other voltage sources are short circuited. This
theorem is valid for linear systems.
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Answer: c
Explanation: In Superposition theorem, while considering a
source, all other current sources are open circuited.
Superposition theorem is not valid for power responses. It is
applicable only for computing voltage and current
responses.
a) 4
b) 5
c) 6
d) 7
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a) 0
b) 1
c) 2
d) 3
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Answer: b
Explanation: The algebraic sum of all the voltages obtained
by considering individual sources is the voltage across 2Ω
resistor. V = 0.97-2.92-1.46 = -3.41V.
a) 1
b) 1.5
c) 2
d) 2.5
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(V-10)/10+V/20+V/2=0 =>
V=1.5V.
10. Find the voltage across 2Ω resistor in the circuit shown
above using Superposition theorem.
a) 1
b) 2
c) 3
d) 4
View Answer
Answer: c
Explanation: The voltage across 2Ωv resistor is the
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a) 8
b) 8.5
c) 9
d) 9.5
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Answer: b
Explanation: The thevenin’s voltage is equal to the open
circuit voltage across the terminals AB that is across 12Ω
resistor. Vth = 10×12/14 = 8.57V.
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Answer: c
Explanation: The resistance into the open circuit terminals
is equal to the thevenin’s resistance => Rth = (12×2)/14 =
1.71Ω.
I=8.57/(2.4+1.71)=0.33A.
4. Determine the equivalent thevenin’s voltage between
terminals A and B in the circuit shown below.
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a) 0.333
b) 3.33
c) 33.3
d) 333
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Answer: c
Explanation: Let us find the voltage drop across terminals A
and B. 50-25=10I+5I => I=1.67A. Voltage drop across 10Ω
resistor = 10×1.67=16.7V. So, Vth=VAB=50-
V=50-16.7=33.3V.
Answer: c
Explanation: To find Rth, two voltage sources are removed
and replaced with short circuit. The resistance at terminals
AB then is the parallel combination of the 10Ω resistor and
5Ω resistor => Rth=(10×5)/15=3.33Ω.
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a) 5
b) 15
c) 25
d) 35
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Answer: c
Explanation: Current through 3Ω resistor is 0A. The current
through 6Ω resistor = (50-10)/(10+6)=2.5A. The voltage
drop across 6Ω resistor = 25×6=15V. So the voltage across
terminals A and B = 0+15+10 = 25V.
Answer: d
Explanation: To find Rth, two voltage sources are removed
and replaced with short circuit =>
Rth=(10×6)/(10+6)+3=6.75Ω.
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a) 0.7
b) 1.7
c) 2.7
d) 3.7
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Answer: c
Explanation: The voltage at terminal a is
Va=(100×6)/16=37.5V, The voltage at terminal b is
Vb=(100×8)/23=34.7V. So the voltage across the terminals
ab is Vab=Va-Vb=37.5-34.7=2.7V.
Answer: d
Explanation: To find Rth, two voltage sources are removed
and replaced with short circuit => Rab=(6×10)/(6+10)+
(8×15)/(8+15)=8.96≅9V.
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I=2.7/(8.96+5)=0.193A≅0.2A.
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a) 1
b) 2
c) 3
d) 4
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Answer: d
Explanation: The magnitude of the current in the Norton’s
equivalent circuit is equal to the current passing through the
short circuited terminals that is I=20/5=4A.
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Answer: b
Explanation: Norton’s resistance is equal to the parallel
combination of both the 5Ω and 10Ω resistors that is R =
(5×10)/15 = 3.33Ω.
Answer: b
Explanation: The current passing through the 6Ω resistor
and the voltage across it due to Norton’s equivalent circuit
is I = 4×3.33/(6+3.33) = 1.43A.
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Answer: c
Explanation: The voltage across the 6Ω resistor is V =
1.43×6 = 8.58V. So the current and voltage have same
values both in the original circuit and Norton’s equivalent
circuit.
a) 1
b) 2
c) 3
d) 4
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Answer: d
Explanation: Short circuiting terminals A and B, 20-10(I1)=0,
I1=2A. 10-5(I2), I2=2A. Current flowing through terminals A
and B= 2+2 = 4A.
Answer: c
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Answer: d
Explanation: To solve for Norton’s current we have to find
the current passing through the terminals A and B. Short
circuiting the terminals a and b, I=100/((6×10)/(6+10)+
(15×8)/(15+8))=11.16 ≅ 11A.
Answer: b
Explanation: The resistance at terminals AB is the parallel
combination of the 10Ω resistor and the 6Ω resistor and
parallel combination of the 15Ω resistor and the 8Ω resistor
=> R=(10×6)/(10+6)+(15×8)/(15+8)=8.96≅9Ω.
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above.
a) 7
b) 8
c) 9
d) 10
View Answer
Answer: a
Explanation: To solve for Norton’s current we have to find
the current passing through the terminals A and B. Short
circuiting the terminals a and b I=11.16×8.96/(5+8.96) =
7.16A.
Answer: d
Explanation: The voltage drop across 5Ω resistor in the
circuit is the product of current and resistance => V =
5×7.16 = 35.8 ≅ 36V.
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Answer: a
Explanation: For the Reciprocity Theorem to satisfy the ratio
of response to the excitation of the circuit should be equal
to the ratio of response to excitation after the source is
replaced.
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b) same
c) before source is replaced is greater than after the source
is replaced
d) before source is replaced is less than after the source is
replaced
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Answer: b
Explanation: For the Reciprocity Theorem to satisfy the ratio
of response to excitation before and after the source is
replaced should be same and if that condition satisfies the
reciprocity theorem is valid for the given circuit.
Answer: c
Explanation: The circuit which satisfies Reciprocity
Theorem is called linear circuit. A linear circuit is an
electronic circuit in which, for a sinusoidal input voltage of
frequency f, any steady-state output of the circuit (the
current through any component, or the voltage between any
two points) is also sinusoidal with frequency f.
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a) 0.143
b) 1.43
c) 14.3
d) 143
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Answer: b
Explanation: Total resistance in the circuit= 2+[3||(2+2│├|2)
]=3.5Ω. The current drawn by the circuit (It)=20/3.5=5.71Ω.
The current drawn by 2Ω resistor = 1.43A.
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Answer: a
Explanation: The ratio of response to excitation before the
source is replaced is equal to 0.0715. And the ratio of
response to excitation before the source is replaced is
equal to 0.0715. So, the circuit satisfies the Reciprocity
theorem.
a) 1
b) 2
c) 3
d) 4
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Answer: b
Explanation: The 6Ω resistor is parallel to 3Ω resistor and
the resultant is in series with 2Ω resistor. Total current from
source = 12/(2+(6│|3) )=3A. Current through 3Ω resistor=3×
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6/(6+3)=2A.
Answer: b
Explanation: The ratio of response to excitation before the
source is replaced is equal to 0.167. And the ratio of
response to excitation before the source is replaced is
equal to 0.167. So, the circuit satisfies the Reciprocity
theorem.
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Answer: c
Explanation: While considering Reciprocity theorem, we
consider ratio of response to excitation as ratio of voltage to
current or current to voltage.
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Answer: c
Explanation: Reciprocity Theorem is applied for linear
bilateral networks, not for linear or for linear bilateral or for
lumped networks.
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d) Power
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Answer: b
Explanation: Reciprocity Theorem is used to find the
change in voltage or current when the resistance is
changed in the circuit. If reciprocity theorem is satisfied the
ratio of response to excitation is same for the two
conditions.
a) 1
b) 2
c) 3
d) 4
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Answer: a
Explanation: Total resistance in the circuit = 2+[3||(2+2||2)]
= 3.5Ω. The total current drawn by the circuit =10/(4+6||3) =
1.67A. Current through 3Ω resistor = 1.11A ≅1A.
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c) 0.93
d) 0.94
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a) 0.33
b) 0.44
c) 0.55
d) 0.66
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Answer: c
Explanation: Total resistance in the circuit = 4+6||3Ω. The
total current drawn by the circuit =10/(4+6||3)=1.67A.
Current through 6Ω resistor = 0.55A.
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a) 0.11
b) 0.22
c) 0.33
d) 0.44
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Answer: c
Explanation: Total current in the circuit = 10/(4+3||2||6)=2A.
Current through 6Ω resistor= 2×(3||2)/(6+3||2)=0.33A.
Answer: c Explanation:
a) 0.45
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b) 0.56
c) 0.67
d) 0.78
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Answer: c
Explanation: Total current = 10 / (4 + (6||2||3) = 2A. Current
through 3Ω resistor= 2 x (6||2) / (3 + (6||2)) =0.67A.
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Answer: c
Explanation: 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.
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Answer: c
Explanation: 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.
Answer: b
Explanation: 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
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d) RL = RS
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Answer: d
Explanation: 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.
Answer: d
Explanation: 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.
a) 15-j20
b) 15+j20
c) 20-j15
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d) 20+j15
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Answer: a
Explanation: The maximum power transfer occurs when the
load impedance is equal to the complex conjugate of
source impedance ZS. ZL = ZS* = (15-j20) Ω.
Answer: b
Explanation: 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.
Answer: c
Explanation: 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.
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a) 1
b) 2
c) 3
d) 4
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Answer: b
Explanation: As RL is fixed, the maximum power transfer
theorem does not apply. Maximum current flows in the
circuit when RS is minimum. So RS = 2Ω.
Answer: c
Explanation: ZT = RS – j5+ RL = 2-j5+20 = 22.56∠-12.8⁰Ω.
I=VS/ZT = -50∠0⁰/22.56∠-12.8⁰ = 2.22∠-12.8⁰A. P = I2R=
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2.222×20=98.6W.
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Answer: a
Explanation: In an electrical circuit itself there are pairs of
terms which can be interchanged to get new circuits. The
dual pair of current is voltage. And the dual pair of voltage
is current.
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Answer: d
Explanation: The dual pair of inductance is capacitance.
And the dual pair of capacitance is inductance.In an
electrical circuit itself there are pairs of terms which can be
interchanged to get new circuits.
Answer: c
Explanation: The dual pair of resistance is conductance.
And the dual pair of conductance is resistance.
Answer: b
Explanation: The dual pair of voltage source is current
source. And the dual pair of current source is voltage
source.
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b) current
c) voltage
d) current source
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Answer: a
Explanation: In an electrical circuit itself there are pairs of
terms which can be interchanged to get new circuits. The
dual pair of KCL is KVL. And the dual pair of KVL is KCL.
Answer: d
Explanation: Tellegen’s Theorem is valid for any lumped
network. So, Tellegan’s theorem is valid for linear or non-
linear networks, passive or active networks and time variant
or time invariant networks.
Answer: c
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a) satisfies
b) does not satisfy
c) satisfies partially
d) satisfies only for some elements
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a) True
b) False
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Answer: b
Explanation: Millman’s Theorem states that if there are
voltage sources V1, V2,…… Vn with internal resistances
R1, R2,…..Rn, respectively, are in parallel, then these
sources are replaced by single voltage source V’ in series
with R’.
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a) V‘=(V1G1+V2G2+⋯.+VnGn)
b) V‘=((V1G1+V2G2+⋯.+VnGn))/((1/G1+1/G2+⋯1/Gn))
c) V‘=((V1G1+V2G2+⋯.+VnGn))/(G1+G2+⋯Gn)
d) V‘=((V1/G1+V2/G2+⋯.+Vn/Gn))/( G1+G2+⋯Gn)
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Answer: c
Explanation: The value of equivalent voltage source is V‘=
((V1G1+V2G2+⋯.+VnGn))/(G1+G2+⋯Gn).
Answer: c
Explanation: Let the equivalent resistance is R’. The value
of equivalent resistance is R’=1/((G1+G2+⋯Gn) ).
Answer: c
Explanation: Millman’s Theorem states that if there are
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Answer: a
Explanation: The value of equivalent current source is
I‘=((I1R1+I2R2+⋯.+InRn))/(R1+R2+⋯Rn).
Answer: c
Explanation: Let the equivalent conductance is G’. The
value of equivalent conductance is G’=1/((R1+R2+⋯Rn) ).
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a) 1
b) 2
c) 3
d) 4
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Answer: c
Explanation: Applying Nodal analysis the voltage V is given
by (10-V)/2+(20-V)/5=V/3. V=8.7V. Now the current through
3Ω resistor in the circuit is I = V/3 = 8.7/3 = 2.9A ≅ 3A.
Answer: b
Explanation: V‘=((V1G1+V2G2))/(G1+G2)=(10(1/2)+20(1
/5))/(1/2+1/5)=12.86V. R’=1/((G1+G2) )=1/(1/2+1/5)=1.43Ω.
Current through 3Ω resistor=I=12.86/(3+1.43)=2.9A≅3A.
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a) 2
b) 1.5
c) 1
d) 0.5
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Answer: b
Explanation: Applying Nodal analysis the voltage V is given
by (5-V)/1+(10-V)/3=V/4. V=6V. The current through 4Ω
resistor I = V/4 = 6/4 = 1.5A.
Answer: c
Explanation: V‘=((V1G1+V2G2))/(G1+G2)=(5(1/1)+10(1
/3))/(1/1+1/3)=6.25V. R’=1/((G1+G2) )=1/(1/1+1/3)=0.75Ω.
I=6.25/(4+0.75)=1.5A.
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