Superposition Theorem MCQ Quiz in বাংলা - Objective Question with Answer for Superposition Theorem - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Superposition theorem:

The superposition theorem states that the response of any circuit with multiple voltage and current sources is equal to the algebraic sum of the response produced by individual sources.

Calculation:

Case 1: When 5 V is active:

Applying KVL in loop ABEA:

-5 + 2I₂ + 6I = 0

2I2 + 6I = 5........(i)

Applying KVL in loop ABCEA:

-5 + 2I2 + 4I2 - 4I  = 0

6I2 - 4I = 5..........(ii)

Solving (i) and (ii), we get:

I2 = 1.13 A

I = 0.454 A

Case 2: When 10 V is active:

Applying KVL in loop ABEA:

-10 + 4I2 + 6I = 0

4I2 + 6I = 10..........(i)

Applying KVL in loop ABCEA:

-10 + 4I2 + 2I2 - 2I  = 0

6I2 - 2I = 10..........(ii)

Solving (i) and (ii), we get:

I2 = 1.81 A

I = 0.454 A

I_net = 0.454 + 0.454

Inet = 0.908 A

Superposition theorem requires as many circuits to be solved as there are sources. While using the superposition theorem we consider one source at a time by deactivating all the other sources.

Additional Information 

 Superposition Theorem:

• It is stated that in any linear, active, bilateral network having more than one source, the response across any element is the Algebraic sum of the response obtained from each source considered separately and all other sources are replaced by their internal resistance.
• The principle of the superposition theorem is based on Linearity.
• Voltage Source  →   short
• Current source   →  open
• Do not disturb the dependent source present in the network.

Step to solving Network by superposition theorem

• Step 1 – Take only one independent source of voltage or current and deactivate the other sources.
• Step 2 – If there is a voltage source then short circuit it and if there is a current source then just open-circuit it.
• Step 3 – Thus, by activating one source and deactivating the other source find the current in each branch of the network.
• Step 4 – Now to determine the net branch current utilizing the superposition theorem, add the currents obtained from each individual source for each branch.
• Step 5 – If the current obtained by each branch is in the same direction then add them and if it is in the opposite direction, subtract them to obtain the net current in each branch.

Superposition Theorem:

• It is stated that in any linear, active, bilateral network having more than one source, the response across any element is the Algebraic sum of the response obtained from each source considered separately and all other sources are replaced by their internal resistance.
• The principle of the superposition theorem is based on linearity and homogeneity.
• Voltage Source  →   short
• Current source   →  open
• Do not disturb the dependent source present in the network.

Limitations of SPT:

• This theorem cannot be used to measure power.
• This theorem is not applicable to unbalanced bridge circuits.
• Applicable only to linear circuits not for nonlinear circuits.
• Applicable only for the circuits having more than one source.

Application of SPT:

Superposition Theorem is applied to determine the current in one particular branch of a network containing several voltage source and/or current source.

The Correct Answer is an option (1) 

Concept:

There is more than one source in the above circuit, so we can implement the superposition theorem to analyze the circuit.

Superposition Theorem:

• The superposition theorem states that a circuit with multiple voltage and current sources is equal to the sum of simplified circuits using just one of the sources.
• While implementing the superposition theorem one of the sources is considered and other sources are replaced by their internal resistances.
• Internal Resistance for an ideal voltage source is zero, therefore circuit is short-circuited.
• Internal Resistance for the ideal current source is infinite, therefore circuit is open-circuited.

Now, in the given circuit we need to find the current through 10 Ω, only due to the current source.

Hence, using the superposition theorem considers the current source in the circuit and replaces the voltage source with its internal resistance.

The circuit will be:

From the above circuit, by using the current division rule we can find the current through 10 Ω resistor,

\(I_{10Ω} =2\times \frac{R_1}{R_1+R_2}\)

\(I_{10Ω} =2\times \frac{5}{5+10}\)

\(I_{10Ω} =\frac{2}{3}\ \)

\(I_{10Ω} =0.66\ A\)

Additional InformationWe can obtain the Total current across 10 Ω Resistor by adding current due to both the sources algebraically.

Considering the Voltage source in the circuit and replacing the current source with its internal resistance,

The circuit will be replaced as:

From the above circuit, by using ohm's law we can obtain the current through 10 Ω resistor,

\(I'_{10Ω }=\frac{V}{R_1+R_2}\)

\(I'_{10Ω }=\frac{20}{5+10}\)

\(I'_{10Ω }=1.33A\ \)

Total Current (I_T) = I'_10Ω + I_10Ω

Both currents are flowing opposite each other,

∴ I_T = 1.33 - 0.66

I_T = 0.66 A

In Conclusion, the Overall current will be 0.66 A flows through 10 Ω resistor in the clockwise direction.

Here, we cannot use superposition theorem to calculate the net power as power is a non-linear quantity

Since P = I²R

\(I=\pm \sqrt{\frac{P}{R}}\)

When only E₁ is acting alone, the power consumed by the resistance R is 18 W, i.e.

\(18=I_{1}^{2}R\)

\({{I}_{1}}=\pm \sqrt{\frac{18}{R}}\)

Similarly, we can write

\({{I}_{2}}=\pm \sqrt{\frac{50}{R}}\)

\({{I}_{3}}=\pm \sqrt{\frac{98}{R}}\)

When all the sources are acting simultaneously the net current through the resistance R using Superposition theorem is:

I = ± I₁ ± I₂ ± I₃

\(I=\pm \sqrt{\frac{18}{R}}\pm \sqrt{\frac{50}{R}}\pm \sqrt{\frac{98}{R}}\)

Net power when all the sources are acting simultaneously will be:

P_net = I²R

\({{P}_{net}}={{\left( \pm \sqrt{\frac{18}{R}}\pm \sqrt{\frac{50}{R}}\pm \sqrt{\frac{98}{R}} \right)}^{2}}R\)

\({{P}_{net}}={{\left( \pm \sqrt{18}\pm \sqrt{50}\pm \sqrt{98} \right)}^{2}}\)

\({{P}_{net}}={{\left( \pm 3\sqrt{2}\pm 5\sqrt{2}\pm 7\sqrt{2} \right)}^{2}}\)

The maximum value of the power will happen when all the current will be contributing in the same direction:

i.e. \({{P}_{max}}={{\left( 3\sqrt{2}+5\sqrt{2}+7\sqrt{2} \right)}^{2}}\)

\({{P}_{max}}={{\left( 15\sqrt{2} \right)}^{2}}\)

P_max = 450 W

For minimum power consumed by R, the current contribution by I₁ and I₂ must be same, but opposite to the current contribution by I₃, i.e.

\({{P}_{min}}={{\left( +3\sqrt{2}+5\sqrt{2}-7\sqrt{2} \right)}^{2}}\)

\(or~{{P}_{min}}={{\left( -3\sqrt{2}-5\sqrt{2}+7\sqrt{2} \right)}^{2}}\)

\({{P}_{min}}={{\left( \sqrt{2} \right)}^{2}}\)

The correct answer is 1/15, 4/30, -1/30

Solution:

Apply superposition theorem apply one source at a time 

Case:1 Apply I₁ current source and short circuit voltage source and open circuit I₂

To find the current through the 12Ω resistance apply the current division rule 

\(A I_1 = \frac{I_1}{5} \times \frac{6}{6 + 12}\) = \(\frac{1}{15}\)I₁  

∴ A = \(\frac{1}{15}\)

Case2: Apply I₂ current source and short circuit the voltage source and open circuit the I₁ current source.

Appy current division to find the current through the 12 Ω resistance.

\(B I_2 =[ \frac{4 I_2}{10} \times \frac{6}{6 + 12}] = \frac{4}{30} I_2\)

∴ B = \(\frac{4}{30}\)

Case3: Apply Voltage source and open circuit both the current source.

The current supplied by the voltage source is 

\(\frac{V}{6 + (12 || 6)} = \frac{V}{10}\)

∴ current through 12 Ω resistance is calculated by using the current division law

\(\frac{V}{10} \times \frac{6}{6 + 12}\)

∴ C V =  \(\frac{-1}{30}\)V ( Negative sign is used because the direction of current is opposite to that of in given question)

∴ C = \(\frac{-1}{30}\)

Concept:

Superposition Theorem

• The superposition theorem states that "in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently."
• The superposition theorem applies only when all the components of the circuit are linear, which is the case for resistors, capacitors, and inductors it is not applicable to networks containing nonlinear elements.

Important Notes:

Various Theorem and the circuits where they are applicable are shown below in the table:

Let I1 be current through resistor R due to voltage source V.

\(I_1^2R = 36\;watt\)

\(I_1=\frac{6}{\sqrt R}\)

Similarly I2 be current through resistor R due to the current source I.

\( I_2^2R = 16\;watt\)

\(I_2=\frac{4}{\sqrt R}\)

When both sources are used, by super position theorem:

I = I1 + I2

Hence power dissipated is:

P = I2R = (I1 + I2)2 R

\( P= {\left( {\frac{6}{{\sqrt R }} + \frac{4}{{\sqrt R }}} \right)^2}R\)

= (6 + 4)2

Superposition Theorem:

• The superposition theorem states that "in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently."
• The superposition theorem applies only when all the components of the circuit are linear, which is the case for resistors, capacitors, and inductors it is not applicable to networks containing nonlinear elements.
  Additional Information

​ Superposition Theorem:

• It is stated that in any linear, active, bilateral network having more than one source, the response across any element is the Algebraic sum of the response obtained from each source considered separately and all other sources are replaced by their internal resistance.
• The principle of the superposition theorem is based on Linearity.
• Voltage Source  →   short
• Current source   →  open
• Do not disturb the dependent source present in the network.
•  

  Step to solving Network by superposition theorem

• Step 1 – Take only one independent source of voltage or current and deactivate the other sources.
• Step 2 – If there is a voltage source then short circuit it and if there is a current source then just open-circuit it.
• Step 3 – Thus, by activating one source and deactivating the other source find the current in each branch of the network.
• Step 4 – Now to determine the net branch current utilizing the superposition theorem, add the currents obtained from each individual source for each branch.
• Step 5 – If the current obtained by each branch is in the same direction then add them and if it is in the opposite direction, subtract them to obtain the net current in each branch.