Question 501
The work done during one complete cycle of magnetization is proportional to the:
Options:
A) Length of the magnetic path
B) Area of the hysteresis loop
C) Flux density only
D) Magnetizing current only
Answer: B) Area of the hysteresis loop
Step-by-Step Solution:
The energy lost during one complete cycle of magnetization is represented by the area enclosed by the B-H hysteresis loop.
A larger area means greater energy loss.
Important Notes:
- Energy per cycle ∝ Area of hysteresis loop.
- Unit volume energy equals loop area.
✔ Answer: B) Area of the hysteresis loop
Question 502
The work done per unit volume during one complete cycle of magnetization is equal to the:
Options:
A) Maximum flux density
B) Area of the hysteresis loop
C) Coercive force
D) Retentivity
Answer: B) Area of the hysteresis loop
Step-by-Step Solution:
The area enclosed by the hysteresis loop represents the energy lost per unit volume during one complete cycle.
Therefore, the work done per unit volume equals the loop area.
Important Notes:
- Unit: Joule/m³
- Larger loop → Greater energy loss.
✔ Answer: B) Area of the hysteresis loop
Question 503
The hysteresis loss per second increases with:
Options:
A) Decreasing frequency
B) Increasing frequency
C) Constant frequency only
D) Zero frequency
Answer: B) Increasing frequency
Step-by-Step Solution:
Hysteresis loss is directly proportional to the frequency of magnetization.
According to Steinmetz's equation,
Ph ∝ f
Therefore, increasing the frequency increases hysteresis loss.
Important Notes:
- Higher frequency → Higher hysteresis loss.
- Lower frequency → Lower hysteresis loss.
✔ Answer: B) Increasing frequency
Question 504
According to Steinmetz's equation, hysteresis loss depends on:
Options:
A) Maximum flux density
B) Frequency
C) Volume of magnetic material
D) All of these
Answer: D) All of these
Step-by-Step Solution:
Steinmetz's equation for hysteresis loss is:
Ph = η Bmax^1.6 f V
Hence hysteresis loss depends on:
- Maximum flux density (Bmax)
- Frequency (f)
- Volume of magnetic material (V)
Therefore, all the given factors affect hysteresis loss.
Important Notes:
Steinmetz Equation:
Ph = η Bmax^1.6 f V
where:
- η = Steinmetz constant
- Bmax = Maximum flux density
- f = Frequency
- V = Volume of magnetic material
✔ Answer: D) All of these
Question 505
The constant used in the hysteresis loss equation is known as:
Options:
A) Eddy current constant
B) Steinmetz constant
C) Permeability constant
D) Reluctivity constant
Answer: B) Steinmetz constant
Step-by-Step Solution:
In Steinmetz's hysteresis loss equation,
Ph = η Bmax^1.6 f V
the constant η depends upon the nature of the magnetic material and is called the Steinmetz constant.
Important Notes:
- η is also called the hysteresis constant.
- It depends on the magnetic material.
✔ Answer: B) Steinmetz constant
Question 506
A magnetic circuit having two or more paths for magnetic flux is called a:
Options:
A) Series magnetic circuit
B) Parallel magnetic circuit
C) Closed magnetic circuit
D) Open magnetic circuit
Answer: B) Parallel magnetic circuit
Step-by-Step Solution:
A parallel magnetic circuit provides two or more paths through which magnetic flux can flow.
Its behavior is similar to that of a parallel electric circuit, where current divides into different branches.
Therefore, such a magnetic circuit is called a Parallel Magnetic Circuit.
Important Notes:
- Contains two or more flux paths.
- Magnetic flux divides among the parallel paths.
- Similar to a parallel electric circuit.
✔ Answer: B) Parallel magnetic circuit
Question 507
The behavior of a parallel magnetic circuit is comparable to a:
Options:
A) Series electric circuit
B) Parallel electric circuit
C) Open circuit
D) Short circuit
Answer: B) Parallel electric circuit
Step-by-Step Solution:
Just as electric current divides into different branches in a parallel electric circuit, magnetic flux divides into different magnetic paths in a parallel magnetic circuit.
Hence, the behavior is similar to a parallel electric circuit.
Important Notes:
- Current divides in a parallel electric circuit.
- Flux divides in a parallel magnetic circuit.
✔ Answer: B) Parallel electric circuit
Question 508
In the given parallel magnetic circuit, the current-carrying coil is wound on the:
Options:
A) Left limb
B) Right limb
C) Central limb AB
D) Upper yoke
Answer: C) Central limb AB
Step-by-Step Solution:
The exciting coil is wound on the central limb AB.
When current flows through the coil, magnetic flux is produced in the central limb before splitting into parallel paths.
Important Notes:
- Coil is wound on the central limb.
- Central limb carries the total magnetic flux.
✔ Answer: C) Central limb AB
Question 509
The magnetic flux produced in the central limb of the parallel magnetic circuit is denoted by:
Options:
A) φ₂
B) φ₃
C) φ₁
D) φ₀
Answer: C) φ₁
Step-by-Step Solution:
The exciting coil produces the total magnetic flux in the central limb.
This total flux is represented by φ₁.
It later divides into φ₂ and φ₃.
Important Notes:
- φ₁ = Total flux.
- Produced by the exciting coil.
✔ Answer: C) φ₁
Question 510
The total magnetic flux in a parallel magnetic circuit is equal to:
Options:
A) φ₁ = φ₂ − φ₃
B) φ₁ = φ₂ × Ï†₃
C) φ₁ = φ₂ + φ₃
D) φ₁ = φ₂ / φ₃
Answer: C) φ₁ = φ₂ + φ₃
Step-by-Step Solution:
According to the law of conservation of magnetic flux,
The total flux entering a junction equals the sum of the fluxes in the parallel branches.
Therefore,
φ₁ = φ₂ + φ₃
Important Notes:
Total Flux:
φ₁ = φ₂ + φ₃
Similar to current division in electrical circuits.
✔ Answer: C) φ₁ = φ₂ + φ₃
Question 511
In the illustrated magnetic circuit, the parallel paths are:
Options:
A) AB and BC
B) ADCB and AFEB
C) AD and FE
D) AC and BD
Answer: B) ADCB and AFEB
Step-by-Step Solution:
After reaching point A, the magnetic flux divides into two branches:
- ADCB
- AFEB
These two branches form the parallel magnetic circuit.
Important Notes:
Parallel magnetic paths:
- ADCB
- AFEB
✔ Answer: B) ADCB and AFEB
Question 512
The reluctance of a magnetic path depends upon its:
Options:
A) Length
B) Cross-sectional area
C) Permeability of the material
D) All of these
Answer: D) All of these
Step-by-Step Solution:
Reluctance is given by
S = l / (μA)
where
- l = Length of magnetic path
- μ = Permeability
- A = Cross-sectional area
Therefore, reluctance depends on all three quantities.
Important Notes:
Reluctance Formula:
S = l / (μA)
Higher permeability or area reduces reluctance.
✔ Answer: D) All of these
Question 513
The reluctance of the central path BA is represented by:
Options:
A) S₂
B) S₃
C) S₁
D) S₄
Answer: C) S₁
Step-by-Step Solution:
The reluctance corresponding to the central limb BA is represented by S₁.
The other two branches have reluctances S₂ and S₃.
Important Notes:
- S₁ → Central limb BA
- S₂ → Path ADCB
- S₃ → Path AFEB
✔ Answer: C) S₁
Question 514
The total magnetomotive force (MMF) required in a parallel magnetic circuit is equal to the:
Options:
A) MMF of the central limb only
B) MMF of any one parallel path
C) Sum of the MMFs of all individual paths
D) Difference of the MMFs of the parallel paths
Answer: C) Sum of the MMFs of all individual paths
Step-by-Step Solution:
The total magnetomotive force required for the magnetic circuit is obtained by adding the MMF required for the central limb and each parallel magnetic path.
Therefore,
Total MMF = MMF(BA) + MMF(ADCB) + MMF(AFEB)
Important Notes:
Total MMF equals the sum of MMFs required by all magnetic sections.
✔ Answer: C) Sum of the MMFs of all individual paths
Question 515
The magnetomotive force (MMF) of a magnetic path is equal to:
Options:
A) Flux × Reluctance
B) Flux ÷ Reluctance
C) Reluctance ÷ Flux
D) Permeability × Flux
Answer: A) Flux × Reluctance
Step-by-Step Solution:
According to Ohm's law for magnetic circuits,
MMF = Flux × Reluctance
That is,
F = ΦS
where
- F = Magnetomotive Force (AT)
- Φ = Magnetic Flux (Wb)
- S = Reluctance (AT/Wb)
Important Notes:
Magnetic Circuit Law:
MMF = Flux × Reluctance
This is analogous to:
Voltage = Current × Resistance in electric circuits.
✔ Answer: A) Flux × Reluctance
Question 516
Mutual Inductance is the property of a coil by which it opposes the change of current in:
Options:
A) The same coil only
B) The neighbouring coil
C) The resistor
D) The capacitor
Answer: B) The neighbouring coil
Step-by-Step Solution:
Mutual inductance is the property by which a changing current in one coil induces an EMF in another nearby coil.
Thus, one coil opposes the change of current occurring in the neighbouring coil.
Therefore, the correct answer is the neighbouring coil.
Important Notes:
- Mutual inductance exists between two coils.
- It is caused by changing magnetic flux.
- Measured in Henry (H).
✔ Answer: B) The neighbouring coil
Question 517
The EMF induced in a neighbouring coil due to changing current in another coil is called:
Options:
A) Self-induced EMF
B) Mutually induced EMF
C) Back EMF
D) Counter EMF
Answer: B) Mutually induced EMF
Step-by-Step Solution:
When current changes in one coil, the changing magnetic flux links the nearby coil and induces an EMF.
This induced EMF is known as mutually induced EMF.
Important Notes:
- Caused by changing magnetic flux.
- Based on Faraday's Law.
- Exists only between nearby coils.
✔ Answer: B) Mutually induced EMF
Question 518
The phenomenon responsible for inducing EMF in a neighbouring coil is known as:
Options:
A) Self Induction
B) Mutual Induction
C) Electrostatic Induction
D) Polarization
Answer: B) Mutual Induction
Step-by-Step Solution:
When changing magnetic flux produced by one coil links another coil, an EMF is induced.
This phenomenon is called Mutual Induction.
Important Notes:
- Requires two coils.
- Requires changing current.
- Used in transformers.
✔ Answer: B) Mutual Induction
Question 519
When the switch connected to Coil A is closed, the magnetic flux produced links with:
Options:
A) Coil A only
B) Coil B only
C) Both Coil A and Coil B
D) Neither coil
Answer: C) Both Coil A and Coil B
Step-by-Step Solution:
Current flowing in Coil A produces magnetic flux.
This flux links both the coil producing it and the nearby coil.
Therefore, both coils are linked by the magnetic flux.
Important Notes:
- Self flux links Coil A.
- Mutual flux links Coil B.
✔ Answer: C) Both Coil A and Coil B
Question 520
Mutual inductance occurs only when the current in the primary coil is:
Options:
A) Constant
B) Zero
C) Changing
D) Maximum
Answer: C) Changing
Step-by-Step Solution:
According to Faraday's Law, EMF is induced only when magnetic flux changes.
A changing current changes the magnetic flux.
Hence, mutual inductance occurs only when current is changing.
Important Notes:
- Constant current produces no induced EMF.
- Changing current produces changing flux.
✔ Answer: C) Changing
Question 521
The SI unit of Mutual Inductance is:
Options:
A) Weber
B) Tesla
C) Henry
D) Ampere
Answer: C) Henry
Step-by-Step Solution:
Mutual inductance is measured in Henry (H).
One Henry is the mutual inductance when an EMF of 1 volt is induced due to a current change of 1 ampere per second.
Important Notes:
SI Unit = Henry (H)
✔ Answer: C) Henry
Question 522
Two coils are said to have a mutual inductance of one Henry when:
Options:
A) One ampere flows continuously.
B) One volt is induced when the current changes at one ampere per second.
C) One volt is applied continuously.
D) Flux becomes one Weber.
Answer: B) One volt is induced when the current changes at one ampere per second.
Step-by-Step Solution:
By definition,
If
- Induced EMF = 1 V
- Rate of change of current = 1 A/s
then the mutual inductance is 1 Henry.
Important Notes:
Definition of 1 Henry:
- EMF = 1 Volt
- Current change = 1 A/s
✔ Answer: B) One volt is induced when the current changes at one ampere per second.
Question 523
Mutual inductance depends upon the:
Options:
A) Number of turns
B) Cross-sectional area
C) Closeness of the coils
D) All of these
Answer: D) All of these
Step-by-Step Solution:
Mutual inductance increases with:
- Number of turns
- Cross-sectional area
- Closeness between coils
Therefore, all factors affect mutual inductance.
Important Notes:
Factors affecting Mutual Inductance:
- Number of turns
- Area
- Distance between coils
✔ Answer: D) All of these
Question 524
Mutual inductance increases when the two coils are placed:
Options:
A) Far apart
B) Very close together
C) At right angles
D) In different rooms
Answer: B) Very close together
Step-by-Step Solution:
The closer the coils are, the greater the magnetic flux linkage.
Hence, mutual inductance increases.
Important Notes:
Closer coils → Higher flux linkage → Higher mutual inductance.
✔ Answer: B) Very close together
Question 525
A transformer works on the principle of:
Options:
A) Self Induction
B) Mutual Induction
C) Electrostatic Induction
D) Electromagnetic Attraction
Answer: B) Mutual Induction
Step-by-Step Solution:
A transformer transfers electrical energy from the primary winding to the secondary winding through mutual induction.
The changing current in the primary produces changing magnetic flux, inducing EMF in the secondary.
Important Notes:
- Transformer works on Mutual Induction.
- Based on Faraday's Law.
- Requires alternating current (AC).
✔ Answer: B) Mutual Induction
Question 526
Find the values of the mesh currents , , and flowing clockwise in the first, second, and third meshes respectively.
Options:
A) 1.54 A, −0.189 A, −1.195 A
B) 2.34 A, −3.53 A, −2.23 A
C) 4.33 A, 0.55 A, 6.02 A
D) −1.18 A, −1.17 A, −1.16 A
Answer: A) 1.54 A, −0.189 A, −1.195 A
Step-by-Step Solution:
Assume clockwise mesh currents as:
- I₁ → Left mesh
- I₂ → Middle mesh
- I₃ → Right mesh
Apply Kirchhoff's Voltage Law (KVL) to each mesh.
Mesh 1:
Across the 1 Ω resistor and the shared 2 Ω resistor,
1I₁ + 2(I₁ − I₂) = 5
⇒ 3I₁ − 2I₂ = 5
or
−3I₁ + 2I₂ − 5 = 0
Mesh 2:
Across the shared 2 Ω resistor, 3 Ω resistor, and shared 4 Ω resistor,
2(I₂ − I₁) + 3I₂ + 4(I₂ − I₃) = 0
⇒ −2I₁ + 9I₂ − 4I₃ = 0
or
2I₁ − 9I₂ + 4I₃ = 0
Mesh 3:
Across the shared 4 Ω resistor and the 5 Ω resistor,
4(I₃ − I₂) + 5I₃ = −10
⇒ −4I₂ + 9I₃ = −10
or
4I₂ − 9I₃ − 10 = 0
The three mesh equations are:
−3I₁ + 2I₂ − 5 = 0
2I₁ − 9I₂ + 4I₃ = 0
4I₂ − 9I₃ − 10 = 0
Solving these simultaneous equations,
I₁ = 1.54 A
I₂ = −0.189 A
I₃ = −1.195 A
The negative values of I₂ and I₃ indicate that their actual directions are opposite to the assumed clockwise directions.
Important Notes:
- Apply KVL independently to each mesh.
- For a resistor shared between two meshes, the voltage drop is R(I₁ − I₂).
- A negative mesh current means the actual current flows opposite to the assumed direction.
- Simultaneous equations can be solved using substitution, elimination, or matrix methods.
✔ Answer: A) 1.54 A, −0.189 A, −1.195 A
Question 527
The circuit shown below has a 10 V source on the left, a 15 V source on the right, a 2 Ω resistor, a 4 Ω resistor, and a 3 Ω resistor connected as shown. Find the current flowing through the 2 Ω resistor, 4 Ω resistor, and 3 Ω resistor, respectively.
Options:
A) 0.962 A, 1.731 A, 2.692 A
B) 1.731 A, 0.962 A, 2.692 A
C) 2.692 A, 1.731 A, 0.962 A
D) 0.962 A, 2.692 A, 1.731 A
Answer
✔ A) 0.962 A, 1.731 A, 2.692 A
Step-by-Step Solution
Let the voltage at the common junction of the three resistors be V volts.
Using Kirchhoff's Current Law (KCL),
Current through the 2 Ω resistor,
I₁ = (10 − V) / 2
Current through the 4 Ω resistor,
I₂ = (15 − V) / 4
Current through the 3 Ω resistor,
I₃ = V / 3
According to KCL,
I₁ + I₂ = I₃
Substituting the values,
(10 − V)/2 + (15 − V)/4 = V/3
Taking the LCM (12),
6(10 − V) + 3(15 − V) = 4V
60 − 6V + 45 − 3V = 4V
105 − 9V = 4V
13V = 105
V = 105/13
V = 8.077 V
Now calculate each current.
Current through the 2 Ω resistor
I₁ = (10 − 8.077)/2
I₁ = 1.923/2
I₁ = 0.962 A
Current through the 4 Ω resistor
I₂ = (15 − 8.077)/4
I₂ = 6.923/4
I₂ = 1.731 A
Current through the 3 Ω resistor
I₃ = 8.077/3
I₃ = 2.692 A
Verification:
0.962 + 1.731 = 2.693 A ≈ 2.692 A
Hence, the answer is verified.
Important Points
• Kirchhoff's Current Law (KCL): The sum of currents entering a junction is equal to the sum of currents leaving the junction.
• Ohm's Law:
I = V / R
• First determine the node voltage, then calculate the current through each branch.
• Always verify the result using KCL.
• Small differences in the final answer are due to rounding.
Final Answer
✔ A) 0.962 A, 1.731 A, 2.692 A
Question 528
For the circuit shown below, determine the value of the voltage source V such that no current flows through the 3 Ω resistor.
Options:
A) 6.5 V
B) 7.5 V
C) 8.5 V
D) 10 V
Answer
✔ B) 7.5 V
Step-by-Step Solution
If no current flows through the 3 Ω resistor, then the voltage across it must be zero.
Therefore, the voltage at the left end of the 3 Ω resistor must be equal to the voltage at the right end.
Step 1: Find the voltage at the left junction
The left side consists of a 5 V source, 1 Ω resistor, and 2 Ω resistor.
Using the voltage divider rule,
Left junction voltage
= 5 × 2 / (1 + 2)
= 10 / 3
= 3.33 V
Step 2: Find the voltage at the right junction
The right side consists of an unknown voltage source V, 5 Ω resistor, and 4 Ω resistor.
Again, using the voltage divider rule,
Right junction voltage
= V × 4 / (5 + 4)
= 4V/9
Step 3: Since current through the 3 Ω resistor is zero,
Left junction voltage = Right junction voltage
3.33 = 4V/9
Multiplying both sides by 9,
29.97 = 4V
V = 29.97 / 4
V ≈ 7.5 V
Hence, the required source voltage is
V = 7.5 V
Important Points
• If the current through a resistor is zero, the voltage across that resistor is also zero.
• Voltage Divider Rule:
Vout = Vin × R₂ / (R₁ + R₂)
• Equal node voltages mean no potential difference exists between the nodes.
• When there is no potential difference across a resistor, no current flows through it.
Final Answer
✔ B) 7.5 V
Question 529
For the circuit shown below, determine the value of the source voltage V₁ such that the current through the 1 Ω resistor is zero.
Options:
A) 83.33 V
B) 78.89 V
C) 87.87 V
D) 33.33 V
Answer
✔ A) 83.33 V
Step-by-Step Solution
Since the current through the 1 Ω resistor is zero, no current flows through that branch.
Therefore, the voltage drop across the 1 Ω resistor is zero, and the voltage at both ends of the resistor must be equal.
Step 1: Find the voltage at the right junction
As no current flows through the 1 Ω resistor, there is no voltage drop across it.
Hence,
Right junction voltage = 10 V
Current through the 3 Ω resistor is
I = 10 / 3 = 3.33 A
Step 2: Apply KCL at the middle junction
Since the 1 Ω branch carries zero current,
Current through the 2 Ω resistor = Current through the 3 Ω resistor
Therefore,
I = 10 / 3 A
Voltage drop across the 2 Ω resistor
= (10/3) × 2
= 20/3 V
Hence, the left junction voltage is
10 + 20/3
= 50/3 V
Step 3: Current through the 5 Ω resistor
I = (50/3) / 5
= 10/3 A
Step 4: Current through the 10 Ω resistor
Current entering the left junction is
= Current through 2 Ω + Current through 5 Ω
= 10/3 + 10/3
= 20/3 A
Voltage drop across the 10 Ω resistor
= (20/3) × 10
= 200/3 V
Therefore,
V₁ = Left junction voltage + Voltage drop across 10 Ω
V₁ = 50/3 + 200/3
= 250/3
= 83.33 V
Hence,
V₁ = 83.33 V
Important Points
• If the current through a resistor is zero, the voltage across that resistor is also zero.
• Use Kirchhoff's Current Law (KCL) at every junction.
• Use Ohm's Law:
V = IR
• The sum of currents entering a node equals the sum of currents leaving the node.
• When two points are at the same potential, no current flows through the resistor connecting them.
Final Answer
✔ A) 83.33 V
Question 530
For the circuit shown below, determine the current flowing through the 5 Ω resistor and the 3 Ω resistor.
Options:
A) 1.75 A through 5 Ω, 1.25 A through 3 Ω
B) 0.50 A through 5 Ω, 2.50 A through 3 Ω
C) 2.30 A through 5 Ω, 0.70 A through 3 Ω
D) 3.00 A through 5 Ω, 0 A through 3 Ω
Answer
✔ A) 1.75 A through 5 Ω, 1.25 A through 3 Ω
Step-by-Step Solution
The circuit contains a 3 A current source between the two meshes.
Hence, Supermesh Analysis is used.
Let,
- I₁ = Clockwise mesh current in the left loop.
- I₂ = Clockwise mesh current in the right loop.
Step 1: Write the current source equation
Since the current source lies between the two meshes,
I₂ − I₁ = 3
Step 2: Write the Supermesh equation
Applying KVL around the outer loop,
−5I₁ − 3I₂ = 5
or
5I₁ + 3I₂ = −5
Step 3: Solve the equations
From the current source equation,
I₂ = I₁ + 3
Substitute into the Supermesh equation,
5I₁ + 3(I₁ + 3) = −5
5I₁ + 3I₁ + 9 = −5
8I₁ = −14
I₁ = −1.75 A
Now,
I₂ = −1.75 + 3 = 1.25 A
The negative sign indicates that the actual direction of I₁ is opposite to the assumed clockwise direction.
Therefore, the actual currents are:
- Current through the 5 Ω resistor = 1.75 A
- Current through the 3 Ω resistor = 1.25 A
Important Points
- A current source common to two meshes forms a Supermesh.
- Write one equation for the current source and another using KVL around the outer loop.
- A negative mesh current indicates that the actual current flows opposite to the assumed direction.
- Supermesh analysis simplifies circuits containing current sources between adjacent meshes.
Final Answer
✔ A) 1.75 A through 5 Ω resistor, 1.25 A through 3 Ω resistor
Question 531
For the circuit shown below, I₁ is the current flowing in the left mesh, I₂ is the current flowing in the right mesh, and I₃ is the current flowing in the top mesh. If all the mesh currents are assumed to flow in the clockwise direction, find the values of I₁, I₂, and I₃.
Options:
A) 7.67 A, 10.67 A, 2 A
B) 10.67 A, 7.67 A, 2 A
C) 7.67 A, 8.67 A, 2 A
D) 3.67 A, 6.67 A, 2 A
Answer
✔ A) 7.67 A, 10.67 A, 2 A
Step-by-Step Solution
The circuit contains a 2 A current source in the top mesh and a 3 A current source common to the left and right meshes.
Step 1: Write the mesh current equations
Since the top mesh contains an independent current source,
I₃ = 2 A
The 3 A current source lies between the left and right meshes, forming a supermesh.
Hence,
I₂ − I₁ = 3
Applying Kirchhoff's Voltage Law (KVL) around the supermesh,
−2I₁ − I₂ = −26
Step 2: Solve the equations
From
I₂ = I₁ + 3
Substitute into the KVL equation,
−2I₁ − (I₁ + 3) = −26
−3I₁ − 3 = −26
−3I₁ = −23
I₁ = 23/3 = 7.67 A
Now,
I₂ = 7.67 + 3
I₂ = 10.67 A
From the first equation,
I₃ = 2 A
Therefore,
I₁ = 7.67 A
I₂ = 10.67 A
I₃ = 2 A
Important Points
- A current source present only in one mesh directly determines that mesh current.
- A current source common to two meshes forms a supermesh.
- Apply Kirchhoff's Voltage Law (KVL) around the supermesh.
- Use the current source equation together with the KVL equation to determine the mesh currents.
- Mesh analysis is applicable only to planar circuits.
Final Answer
✔ A) 7.67 A, 10.67 A, 2 A
Question 532
For the circuit shown below, determine the mesh currents I₁, I₂, and I₃. The meshes containing the 3 A current source form a supermesh.
Options:
A) 7 A, 6 A, 6.22 A
B) 2 A, 1 A, 0.57 A
C) 3 A, 4 A, 5.88 A
D) 6 A, 7 A, 8.99 A
Answer
✔ B) 2 A, 1 A, 0.57 A
Step-by-Step Solution
Let the clockwise mesh currents be:
- I₁ = Left mesh current
- I₂ = Upper-right mesh current
- I₃ = Lower-right mesh current
Since the 3 A current source lies between the left and lower-right meshes, they form a supermesh.
Step 1: Write the current source equation
The current source gives the relation
I₁ − I₃ = −3
Step 2: Apply KVL to the upper-right mesh
Applying Kirchhoff's Voltage Law,
4I₁ − 14I₂ + 11I₃ = 10
Step 3: Apply KVL to the supermesh
Applying KVL around the outer loop,
4I₁ − 28I₂ + 10I₃ = 0
Step 4: Solve the simultaneous equations
The three equations are
I₁ − I₃ = −3
4I₁ − 14I₂ + 11I₃ = 10
4I₁ − 28I₂ + 10I₃ = 0
Solving these equations,
I₁ = −1 A
I₂ = 0.57 A
I₃ = 2 A
The negative sign for I₁ indicates that the actual current flows opposite to the assumed clockwise direction.
Therefore, the actual mesh currents are
- I₁ = 1 A
- I₂ = 0.57 A
- I₃ = 2 A
Hence, the correct option is
2 A, 1 A, 0.57 A
Important Points
- A current source common to two meshes creates a supermesh.
- Apply Kirchhoff's Voltage Law (KVL) around the supermesh.
- Use the current source equation as an additional constraint.
- A negative mesh current means the actual current flows opposite to the assumed direction.
- Mesh analysis is applicable only to planar circuits.
Final Answer
✔ B) 2 A, 1 A, 0.57 A
Question 533
Mesh analysis employs the method of ___________.
Options:
A) Kirchhoff's Voltage Law (KVL)
B) Kirchhoff's Current Law (KCL)
C) Both KVL and KCL
D) Neither KVL nor KCL
Answer
✔ A) Kirchhoff's Voltage Law (KVL)
Step-by-Step Solution
Mesh analysis is one of the most widely used techniques for analyzing electrical circuits. It is used to determine the unknown currents flowing through the meshes (independent loops) of a planar circuit.
The entire method is based on Kirchhoff's Voltage Law (KVL).
Kirchhoff's Voltage Law states that:
"The algebraic sum of all the voltages around any closed loop in an electrical circuit is always zero."
Mathematically,
ΣV = 0
In mesh analysis, the following procedure is followed:
Step 1: Assign a mesh current to each independent loop.
Step 2: Assume all mesh currents flow in the clockwise direction (or any chosen direction).
Step 3: Apply KVL around each mesh by adding all voltage drops and voltage rises.
Step 4: Solve the simultaneous equations to determine the unknown mesh currents.
Although Ohm's Law (V = IR) is used to express the voltage drops across resistors, the fundamental law employed in mesh analysis is Kirchhoff's Voltage Law (KVL).
Therefore, the correct answer is Kirchhoff's Voltage Law (KVL).
Important Points
• Mesh analysis is based entirely on Kirchhoff's Voltage Law (KVL).
• KVL states that the algebraic sum of voltages around a closed loop is zero.
• Ohm's Law (V = IR) is used only to calculate voltage drops.
• Mesh analysis is mainly used for circuits having voltage sources.
• It is one of the simplest methods for determining unknown mesh currents.
Final Answer
✔ A) Kirchhoff's Voltage Law (KVL)
Question 534
Mesh analysis is generally used to determine __________.
Options:
A) Voltage
B) Current
C) Resistance
D) Power
Answer
✔ B) Current
Step-by-Step Solution
The primary objective of mesh analysis is to calculate the current flowing through each mesh of a planar electrical circuit.
In this method,
- A separate current is assumed in every independent mesh.
- Kirchhoff's Voltage Law (KVL) is applied to each mesh.
- A set of simultaneous equations is obtained.
- Solving these equations gives the value of the mesh currents.
Once the mesh currents are known, other electrical quantities can easily be determined.
For example,
Voltage across a resistor:
V = IR
Power consumed by a resistor:
P = I²R
Thus, mesh analysis directly determines the currents, while voltages and power are calculated from these currents.
Therefore, the correct answer is Current.
Important Points
• Mesh analysis is primarily used to determine mesh currents.
• It is based on Kirchhoff's Voltage Law (KVL).
• Voltage can be calculated using Ohm's Law after finding the current.
• Power can be calculated using
P = VI
or
P = I²R
• Mesh analysis reduces the number of simultaneous equations in many circuit problems.
Final Answer
✔ B) Current
Question 535
Mesh analysis can be used for __________.
Options:
A) Planar circuits
B) Non-planar circuits
C) Both planar and non-planar circuits
D) Neither planar nor non-planar circuits
Answer
✔ A) Planar circuits
Step-by-Step Solution
A planar circuit is a circuit that can be drawn on a flat surface without any conductor crossing another conductor.
Mesh analysis requires clearly defined independent meshes (loops).
In a planar circuit, these meshes are easy to identify, making it possible to apply Kirchhoff's Voltage Law (KVL) around each mesh.
However, in a non-planar circuit, conductors cross each other, making it difficult or impossible to define independent meshes properly.
Therefore, mesh analysis cannot be applied directly to non-planar circuits.
For non-planar circuits, Nodal Analysis is generally preferred.
Hence, mesh analysis is applicable only to planar circuits.
Important Points
• Mesh analysis is applicable only to planar circuits.
• A planar circuit can be drawn without crossing branches.
• A mesh is the smallest independent closed path in a circuit.
• Mesh analysis is based on Kirchhoff's Voltage Law (KVL).
• For non-planar circuits, Nodal Analysis is generally more suitable.
• Mesh analysis becomes easier when the number of meshes is less than the number of nodes.
Final Answer
✔ A) Planar circuits
Question 536
For the circuit shown below, determine the node voltage V using the Nodal Analysis method.
Options:
A) −60 V
B) 60 V
C) 40 V
D) −40 V
Answer
✔ A) −60 V
Step-by-Step Solution
Take the bottom conductor as the reference node (Ground).
Let the voltage at the top node be V.
Apply Kirchhoff's Current Law (KCL) at the node.
The currents are:
- Left current source injects 2 A into the node.
- Right current source draws 8 A from the node.
- Current through the 10 Ω resistor is
I = V/10
Applying KCL,
Current entering the node = Current leaving the node
2 = 8 + V/10
Rearranging,
2 − 8 = V/10
−6 = V/10
Multiplying both sides by 10,
V = −60 V
Therefore, the node voltage is
V = −60 V
Important Points
- In Nodal Analysis, one node is selected as the reference (ground) node.
- Apply Kirchhoff's Current Law (KCL) at every unknown node.
- Current entering a node = Current leaving the node.
-
Current through a resistor is calculated using Ohm's Law:
I = V/R
- A negative node voltage indicates that the node is at a lower potential than the reference node.
Final Answer
✔ A) −60 V
Question 537
For the circuit shown below, determine the node voltages V₁ and V₂ using the Nodal Analysis method.
Options:
A) 12 V, 13 V
B) 26.67 V, 11.33 V
C) 11.33 V, 26.67 V
D) 13 V, 12 V
Answer
✔ C) 11.33 V, 26.67 V
Step-by-Step Solution
Take the bottom conductor as the reference node (Ground).
Let the two unknown node voltages be:
- V₁ = Left node voltage
- V₂ = Right node voltage
Apply Kirchhoff's Current Law (KCL) at each node.
Step 1: Apply KCL at Node V₁
Current leaving through the 2 Ω resistor connected to ground:
V₁ / 2
Current leaving through the 2 Ω resistor connecting V₁ and V₂:
(V₁ − V₂) / 2
The 2 A current source enters Node V₁, while the 4 A current source leaves Node V₁ toward Node V₂.
Applying KCL,
V₁/2 + (V₁ − V₂)/2 + 4 = 2
Multiplying throughout by 2,
V₁ + V₁ − V₂ + 8 = 4
2V₁ − V₂ = −4
Step 2: Apply KCL at Node V₂
Current leaving through the 8 Ω resistor:
V₂ / 8
Current leaving through the resistor connecting V₂ and V₁:
(V₂ − V₁) / 2
The 4 A current source enters Node V₂.
The 7 A current source also enters Node V₂.
Applying KCL,
V₂/8 + (V₂ − V₁)/2 = 4 + 7
Multiplying the entire equation by 8,
V₂ + 4(V₂ − V₁) = 88
V₂ + 4V₂ − 4V₁ = 88
−4V₁ + 5V₂ = 88
Step 3: Solve the simultaneous equations
The two nodal equations are
2V₁ − V₂ = −4
−4V₁ + 5V₂ = 88
From the first equation,
V₂ = 2V₁ + 4
Substituting into the second equation,
−4V₁ + 5(2V₁ + 4) = 88
−4V₁ + 10V₁ + 20 = 88
6V₁ = 68
V₁ = 11.33 V
Now,
V₂ = 2(11.33) + 4
V₂ = 26.67 V
Hence,
V₁ = 11.33 V
V₂ = 26.67 V
Important Points
- Select one node as the reference (ground) node.
- Apply Kirchhoff's Current Law (KCL) at every unknown node.
-
Current through a resistor is calculated using Ohm's Law:
I = (V₁ − V₂) / R
- Write one equation for each unknown node voltage.
- Solve the simultaneous equations to obtain the node voltages.
Final Answer
✔ C) 11.33 V, 26.67 V
Question 538
For the circuit shown below, determine the node voltage V using the Nodal Analysis method.
Options:
A) 1 V
B) 2 V
C) 3 V
D) 4 V
Answer
✔ D) 4 V
Step-by-Step Solution
Take the bottom conductor as the reference node (Ground).
Let the voltage at the top node be V.
Apply Kirchhoff's Current Law (KCL) at the node.
Step 1: Determine the current through each branch
Left Branch (10 V source and 2 Ω resistor):
Current flowing from the node to ground is
I₁ = (V − 10) / 2
Middle Branch (1 Ω resistor):
Current is
I₂ = V / 1 = V
Right Branch (7 V source and 3 Ω resistor):
Current is
I₃ = (V − 7) / 3
Step 2: Apply Kirchhoff's Current Law (KCL)
Since there is no current source connected to the node,
Sum of currents leaving the node = 0
(V − 10)/2 + (V − 7)/3 + V = 0
Step 3: Solve the equation
Taking the LCM (6),
3(V − 10) + 2(V − 7) + 6V = 0
Expanding,
3V − 30 + 2V − 14 + 6V = 0
11V − 44 = 0
11V = 44
V = 4 V
Therefore, the node voltage is
V = 4 V
Important Points
- In Nodal Analysis, select one node as the reference (ground) node.
- Apply Kirchhoff's Current Law (KCL) at every unknown node.
-
Current through a resistor is calculated using Ohm's Law:
I = V/R
-
When a voltage source is connected in series with a resistor, the branch current is
I = (Vnode − Vsource) / R
- Solve the nodal equation to determine the unknown node voltage.
Final Answer
✔ D) 4 V
Question 539
For the circuit shown below, determine the node voltages V₁, V₂, and V₃ using the Nodal Analysis
Options:
A) 30.77 V, 7.52 V, 18.82 V
B) 32.34 V, 7.87 V, 8.78 V
C) 34.34 V, 8.99 V, 8.67 V
D) 45.44 V, 6.67 V, 7.77 V
Answer
✔ A) 30.77 V, 7.52 V, 18.82 V
Step-by-Step Solution
Take the bottom conductor as the reference (ground) node.
Let
- V₁ = Left node voltage
- V₂ = Middle node voltage
- V₃ = Right node voltage
Apply Kirchhoff's Current Law (KCL) at each node.
Step 1: Apply KCL at Node V₁
The currents associated with Node V₁ are:
- 8 A current source enters the node.
- 3 A current source leaves the node.
- Current through the 3 Ω resistor = (V₁ − V₂)/3
- Current through the 4 Ω resistor = (V₁ − V₃)/4
Applying KCL,
-8 + (V₁ − V₂)/3 − 3 + (V₁ − V₃)/4 = 0
Step 2: Apply KCL at Node V₂
The currents associated with Node V₂ are:
- 3 A current source enters the node.
- Current through the 1 Ω resistor = V₂
- Current through the 3 Ω resistor = (V₂ − V₁)/3
- Current through the 7 Ω resistor = (V₂ − V₃)/7
Applying KCL,
3 + V₂ + (V₂ − V₃)/7 + (V₂ − V₁)/3 = 0
Step 3: Apply KCL at Node V₃
The currents associated with Node V₃ are:
- 2.5 A current source enters the node.
- Current through the 5 Ω resistor = V₃/5
- Current through the 4 Ω resistor = (V₃ − V₁)/4
- Current through the 7 Ω resistor = (V₃ − V₂)/7
Applying KCL,
-2.5 + (V₃ − V₂)/7 + (V₃ − V₁)/4 + V₃/5 = 0
Step 4: Solve the simultaneous equations
The nodal equations are
-8 + (V₁ − V₂)/3 − 3 + (V₁ − V₃)/4 = 0
3 + V₂ + (V₂ − V₃)/7 + (V₂ − V₁)/3 = 0
-2.5 + (V₃ − V₂)/7 + (V₃ − V₁)/4 + V₃/5 = 0
Solving these equations simultaneously,
V₁ = 30.77 V
V₂ = 7.52 V
V₃ = 18.82 V
Hence, the required node voltages are
V₁ = 30.77 V
V₂ = 7.52 V
V₃ = 18.82 V
Important Points
- Select one node as the reference (ground) node before applying nodal analysis.
- Apply Kirchhoff's Current Law (KCL) at every unknown node.
-
Current through a resistor is calculated using:
I = (V₁ − V₂)/R
- The number of independent nodal equations is equal to the number of unknown node voltages.
- Solve all nodal equations simultaneously to obtain the node voltages.
Final Answer
✔ A) 30.77 V, 7.52 V, 18.82 V
Question 540
For the circuit shown below, determine the node voltages V₁ and V₂ using the Nodal Analysis method.
Options:
A) V₁ = 64.28 V, V₂ = 16.42 V
B) V₁ = 23.32 V, V₂ = 46.45 V
C) V₁ = 87.23 V, V₂ = 29.23 V
D) V₁ = 56.32 V, V₂ = 78.87 V
Answer
✔ A) V₁ = 64.28 V, V₂ = 16.42 V
Step-by-Step Solution
Take the bottom conductor as the reference (ground) node.
Let:
- V₁ = Voltage at the first node
- V₂ = Voltage at the second node
Apply Kirchhoff's Current Law (KCL) at each node.
Step 1: Apply KCL at Node V₁
At node V₁,
- A 12 A current source injects current into the node.
- Current through the 10 Ω resistor = V₁ / 10
-
Current through the 5 Ω resistor connected toward node V₂ is
(V₁ − V₂ + 20) / 5
Applying KCL,
12 = V₁/10 + (V₁ − V₂ + 20)/5
Simplifying,
0.3V₁ − 0.2V₂ = 16
Step 2: Apply KCL at Node V₂
At node V₂,
Current leaves through:
- Left branch toward V₁
- Middle branch through 5 Ω and 15 V source
- Right branch through 5 Ω and 10 V source
Applying KCL,
(V₂ − V₁ − 20)/5 + (V₂ − 15)/5 + (V₂ − 10)/5 = 0
Simplifying,
−V₁ + 3V₂ = −15
Step 3: Solve the Simultaneous Equations
The nodal equations are:
0.3V₁ − 0.2V₂ = 16
−V₁ + 3V₂ = −15
Solving simultaneously,
V₁ = 64.28 V
V₂ = 16.42 V
Therefore,
Node Voltage V₁ = 64.28 V
Node Voltage V₂ = 16.42 V
Important Points
- Choose one node as the reference (ground) before writing KCL equations.
- Apply Kirchhoff's Current Law (KCL) at every unknown node.
-
Current through a resistor is calculated using:
I = (V₁ − V₂) / R
- Include the effect of voltage sources while writing branch current equations.
- Solve the simultaneous equations to obtain the unknown node voltages.
Final Answer
✔ A) V₁ = 64.28 V, V₂ = 16.42 V
Question 541
Nodal analysis is generally used to determine:
Options:
A) Voltage
B) Current
C) Resistance
D) Power
Answer
✔ A) Voltage
Step-by-Step Solution
Nodal Analysis is a circuit analysis technique based on Kirchhoff's Current Law (KCL).
In this method:
- One node is selected as the reference (ground) node.
- The voltages of all other nodes are taken with respect to the reference node.
- KCL equations are written at each unknown node.
- Solving these equations gives the node voltages directly.
Therefore, nodal analysis is primarily used to determine voltages.
Important Points
- Nodal Analysis is based on Kirchhoff's Current Law (KCL).
- It directly calculates node voltages.
- One node is always chosen as the reference (0 V).
- After finding node voltages, branch currents can be calculated using Ohm's Law.
✔ Final Answer: A) Voltage
Question 542
If there are 10 nodes in a circuit, how many independent nodal equations are required?
Options:
A) 10
B) 9
C) 8
D) 7
Answer
✔ B) 9
Step-by-Step Solution
In nodal analysis:
- One node is selected as the reference (ground) node.
- The voltage of the reference node is known (0 V).
- Therefore, KCL equations are written only for the remaining nodes.
Number of independent nodal equations:
Number of Equations = Total Nodes − 1
Given,
Total nodes = 10
Therefore,
Number of equations = 10 − 1 = 9
Hence, 9 independent equations are required.
Important Points
- Number of nodal equations = Total Nodes − 1
- One node is always taken as the reference node (Ground).
- The reference node voltage is 0 V.
- Each remaining node contributes one independent KCL equation.
✔ Final Answer: B) 9
Question 543
Nodal analysis can be applied for:
Options:
A) Planar networks
B) Non-planar networks
C) Both planar and non-planar networks
D) Neither planar nor non-planar networks
Answer
✔ C) Both planar and non-planar networks
Step-by-Step Solution
Unlike mesh analysis, nodal analysis does not depend on the shape of the circuit.
It only requires:
- Identification of circuit nodes.
- Application of Kirchhoff's Current Law (KCL).
Since every electrical circuit has nodes, nodal analysis can be applied to:
- Planar circuits
- Non-planar circuits
Therefore, it is suitable for both types of networks.
Important Points
- Nodal Analysis uses KCL.
- It works for both planar and non-planar circuits.
- It is especially useful when many current sources are present.
- Mesh analysis is generally limited to planar circuits, whereas nodal analysis is not.
✔ Final Answer: C) Both planar and non-planar networks
Question 544
How many reference nodes are selected in nodal analysis?
Options:
A) 1
B) 2
C) 3
D) 4
Answer
✔ A) 1
Step-by-Step Solution
In nodal analysis:
- One node is selected as the reference (ground) node.
- The voltage at this node is assumed to be 0 V.
- All other node voltages are measured with respect to this reference node.
Therefore, only one reference node is selected.
Important Points
- One node is always chosen as the reference (ground) node.
- Reference node voltage = 0 V.
- All unknown node voltages are measured relative to this node.
- Choosing a suitable reference node simplifies circuit calculations.
✔ Final Answer: A) 1
Question 545
In the Superposition Theorem, when considering the effect of one voltage source, all the other voltage sources are:
Options:
A) Shorted
B) Opened
C) Removed
D) Undisturbed
Answer
✔ A) Shorted
Step-by-Step Solution
According to the Superposition Theorem, only one independent source is considered at a time.
When analyzing the effect of one voltage source:
- All other independent voltage sources are replaced by their internal resistance.
- An ideal voltage source has zero internal resistance, so it is replaced by a short circuit.
- Independent current sources are opened.
Therefore, the other voltage sources are short-circuited.
Important Points
- Ideal voltage source → Replace with a short circuit.
- Ideal current source → Replace with an open circuit.
- Only one independent source remains active at a time.
- Superposition applies only to linear circuits.
✔ Final Answer: A) Shorted
Question 546
In the Superposition Theorem, when considering the effect of one current source, all the other voltage sources are:
Options:
A) Shorted
B) Opened
C) Removed
D) Undisturbed
Answer
✔ A) Shorted
Step-by-Step Solution
While applying the Superposition Theorem, only one independent source is kept active.
When considering the effect of one current source:
- All other voltage sources are replaced by their internal resistance.
- Since an ideal voltage source has zero internal resistance, it is replaced by a short circuit.
- Other current sources are opened.
Hence, all other voltage sources are short-circuited.
Important Points
- Voltage sources are always short-circuited while deactivating them.
- Current sources are always open-circuited while deactivating them.
- Only one independent source is active during each step of analysis.
- Final voltage or current is obtained by adding the individual contributions algebraically.
✔ Final Answer: A) Shorted
Question 547
In the Superposition Theorem, when considering the effect of one voltage source, all the other current sources are:
Options:
A) Shorted
B) Opened
C) Removed
D) Undisturbed
Answer
✔ B) Opened
Step-by-Step Solution
According to the Superposition Theorem, all independent sources except one are deactivated.
When analyzing the effect of one voltage source:
- Other voltage sources are replaced by short circuits.
- Other current sources are replaced by their internal resistance.
- Since an ideal current source has infinite internal resistance, it is replaced by an open circuit.
Therefore, all other current sources are opened.
Important Points
- Ideal current source → Open circuit.
- Ideal voltage source → Short circuit.
- Analyze one independent source at a time.
- Sum the individual effects to obtain the final response.
✔ Final Answer: B) Opened
Question 548
In the Superposition Theorem, when considering the effect of one current source, all the other current sources are:
Options:
A) Shorted
B) Opened
C) Removed
D) Undisturbed
Answer
✔ B) Opened
Step-by-Step Solution
When applying the Superposition Theorem, only one independent source is kept active.
If one current source is being considered:
- All other current sources are replaced by their internal resistance.
- An ideal current source has infinite internal resistance.
- Therefore, it is replaced by an open circuit.
- Voltage sources are replaced by short circuits.
Hence, all other current sources are opened.
Important Points
- Ideal voltage source → Short circuit.
- Ideal current source → Open circuit.
- Superposition is valid only for linear electrical networks.
- Calculate the contribution of each source separately and add them algebraically.
✔ Final Answer: B) Opened
Question 549
Find the value of due to the 16 V source using the Superposition Theorem.
Circuit Diagram:
Options:
A) 4.2 V
B) 3.2 V
C) 2.3 V
D) 6.3 V
Answer
✔ B) 3.2 V
Step-by-Step Solution
According to the Superposition Theorem, only one independent source is considered at a time.
Step 1: Keep the 16 V source active
Deactivate all the other independent sources.
- Replace the 10 V voltage source with a short circuit.
- Replace the 15 A current source with an open circuit.
- Replace the 3 A current source with an open circuit.
After deactivating these sources, the circuit reduces to a simple series voltage divider consisting of:
- 20 Ω resistor
- 80 Ω resistor
- 16 V supply
Step 2: Apply the Voltage Divider Rule
The voltage across the 20 Ω resistor is
Vx = Vs × R / (R₁ + R₂)
Substituting the values,
Vx = 16 × 20 / (20 + 80)
Vx = 16 × 20 / 100
Vx = 3.2 V
Therefore,
Vx = 3.2 V
Important Points
- Superposition Theorem is applicable only to linear circuits.
-
While considering one source:
- Voltage sources are replaced by short circuits.
- Current sources are replaced by open circuits.
-
Voltage Divider Rule:
V = Vs × R / (R₁ + R₂)
- The final voltage or current in the circuit is obtained by algebraically adding the contributions of all independent sources.
✔ Final Answer: B) 3.2 V
Question 550
Find the value of due to the 3 A current source using the Superposition Theorem.
Options:
A) 56 V
B) 78 V
C) 38 V
D) 48 V
Answer
✔ D) 48 V
Step-by-Step Solution
According to the Superposition Theorem, only the 3 A current source is kept active.
Step 1: Deactivate the Remaining Independent Sources
- Replace the 16 V voltage source with a short circuit.
- Replace the 10 V voltage source with a short circuit.
- Replace the 15 A current source with an open circuit.
The remaining circuit consists of:
- A 3 A current source
-
Two parallel resistors:
- 20 Ω
- 80 Ω
Step 2: Apply the Current Divider Rule
The current through the 20 Ω resistor is
I₂₀ = I × R(other) / (R₁ + R₂)
Substituting the values,
I₂₀ = 3 × 80 / (20 + 80)
I₂₀ = 3 × 80 / 100
I₂₀ = 2.4 A
Step 3: Calculate the Voltage
Using Ohm's Law,
Vx = I × R
Vx = 2.4 × 20
Vx = 48 V
Therefore,
Vx = 48 V
Important Points
- In the Superposition Theorem, only one independent source is active at a time.
-
While deactivating sources:
- Voltage source → Short Circuit
- Current source → Open Circuit
-
Current Divider Rule:
I₁ = I × R₂ / (R₁ + R₂)
-
Ohm's Law:
V = IR
- The final circuit voltage or current is obtained by algebraically adding the contributions from all independent sources.