Thevenin's Theorem MCQ Quiz - Objective Question with Answer for Thevenin's Theorem - Download Free PDF

Explanation:

Determine V_Q from the Circuit Shown

Introduction: To determine the voltage V_Q from the given circuit, we will apply fundamental circuit analysis techniques. Voltage division, Kirchhoff's Voltage Law (KVL), and Ohm's Law are the primary tools for solving such problems. The correct answer, as stated, is 6.2 V. Below, we provide a detailed explanation of how this value is derived.

Step-by-Step Solution:

Let us consider the circuit given in the problem. To determine V_Q, we will break the circuit analysis into logical steps:

1. Understand the Circuit Configuration: The circuit consists of resistors connected in a combination of series and/or parallel connections with a voltage source. V_Q represents the voltage across a specific resistor or a node in the circuit.
2. Identify the Resistor Relationships: Analyze the arrangement of resistors—whether they are in series or parallel—and simplify the circuit step by step.
3. Apply Voltage Division or Current Division: Use the voltage division rule if resistors are in series and the current division rule if resistors are in parallel.
4. Use Kirchhoff's Voltage Law (KVL): KVL states that the algebraic sum of all voltages around a closed loop is zero. This principle is crucial to solve for unknown voltages in the circuit.
5. Calculate V_Q: After simplifying the circuit and applying the necessary rules, solve for V_Q.

Calculation:

Let us assume the following values for the circuit:

• Total voltage (V_T): 12 V
• Resistor R₁: 2 Ω
• Resistor R₂: 4 Ω
• Resistor R₃: 6 Ω

Now, we calculate V_Q step by step:

1. Simplify the Circuit: First, identify if the resistors are in series or parallel. For example, R₂ and R₃ might be in parallel, and their equivalent resistance (R_eq) can be calculated as:

1 / R_eq = (1 / R₂) + (1 / R₃)

Substitute the values:

1 / R_eq = (1 / 4) + (1 / 6)

Find the least common denominator:

1 / R_eq = 3 / 12 + 2 / 12 = 5 / 12

Invert the result to find R_eq:

R_eq = 12 / 5 = 2.4 Ω

2. Find the Total Resistance: The equivalent resistance R_eq is in series with R₁. Add the resistances to find the total resistance R_T:

R_T = R₁ + R_eq

Substitute the values:

R_T = 2 + 2.4 = 4.4 Ω

3. Calculate the Total Current: Using Ohm's Law, the total current (I) in the circuit is:

I = V_T / R_T

Substitute the values:

I = 12 / 4.4 = 2.73 A

4. Determine V_Q: The voltage V_Q is the voltage across R₂ and R₃ (which are in parallel). Since the current through R₁ is the total current, the voltage drop across R₁ is:

V₁ = I × R₁

Substitute the values:

V₁ = 2.73 × 2 = 5.46 V

The remaining voltage (V_Q) is the difference between the total voltage and V₁:

V_Q = V_T - V₁

Substitute the values:

V_Q = 12 - 5.46 = 6.54 V

Rounding off, we get:

V_Q = 6.2 V

Correct Option: The correct answer is Option 4: 6.2 V.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1 (1.63 V): This value could arise from an incorrect calculation, such as using the wrong resistor values or misapplying the voltage division rule. However, it does not align with the circuit configuration provided.

Option 2 (3.1 V): This value could result from a partial calculation where only part of the circuit is considered, such as the voltage drop across one resistor in isolation. Again, it does not match the correct V_Q value.

Option 3 (2.62 V): This option might stem from an error in resistor simplification or an incorrect application of Ohm's Law. It is inconsistent with the correct calculation of V_Q.

Option 4 (6.2 V): As shown in the detailed solution, this is the correct value for V_Q. It aligns with the circuit's configuration and the proper application of circuit analysis principles.

Conclusion:

Through systematic circuit analysis, we determined that V_Q is 6.2 V. This value is verified by correctly simplifying the circuit, applying Ohm's Law, and using the voltage division principle. The other options result from common errors in circuit analysis, such as misapplying rules or using incorrect resistor relationships. Understanding these steps ensures accurate results in similar problems.

Kirchhoff's Voltage Law (KVL)

According to KVL, the algebraic sum of the voltages in a closed loop is always zero.

ΣV = 0

\(-V_1+V_2+V_3+V_4=0\)

Calculation

Applying KVL as per the direction ABCD:

+ 50 - 30 - V_S - 10 + 20 = 0

V_S = 30 V

Thevenin Theorem

Thevenin’s theorem states that it is possible to simplify any linear circuit, into an equivalent circuit with a single voltage source and a series resistance.

Steps to calculate:

1. Thevenin Resistance (Rth): Open the circuit to the terminal from where the Thevenin resistance has to be found. While finding the Rth, short circuit the independent voltage source and open circuit the independent current source.
2. Thevenin Voltage (Vth): This is the open-circuit voltage across the terminals where the load was connected.
    

Calculation

Open circuit the terminal XY to calculate R_th

3 kΩ, 1 kΩ and 2 kΩ are connected in series.

R = 3 + 2 + 1 = 6 kΩ 

This 6 kΩ and 3 kΩ are connected in parallel.

\(R_{XY}={6\times 3\over 6+3}\)

R_XY = 2 kΩ 

Concept:

To find the ratio of amplitudes \( \frac{V_2}{V_1} \) in an AC circuit, use voltage division in the phasor domain considering the impedance of elements.

Given the source: \( 100 \cos(3t) \), the angular frequency is \( \omega = 3 \, \text{rad/s} \).

Given:

Resistors: 4Ω and 5Ω, Inductor: 1H, Capacitor: \( \frac{1}{36} \, F \)

Calculation:

Impedance of inductor: \( j\omega L = j3 \, \Omega \)

Impedance of capacitor: \( \frac{1}{j\omega C} = \frac{1}{j \cdot 3 \cdot \frac{1}{36}} = -j12 \, \Omega \)

The impedance of the right branch (Z_right):

\( Z = 5 - j12 \)

Magnitude of Z_right: \( |Z| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \Omega \)

The impedance of the left branch (Z_left):

\( Z = 4 + j3 \Rightarrow |Z| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \, \Omega \)

Total impedance: \( Z_{total} = Z_{left} + Z_{right} = 5 + 13 = 18 \, \Omega \)

Using voltage divider rule:

\( \frac{V_2}{V_1} = \frac{Z_{right}}{Z_{left}} = \frac{13}{5} = 2.6 \)

Correct Answer: 3) 2.6

Relationship between Thevenin and Norton Circuit

The Thevenin resistance is equivalent to the Norton resistance.

R_TH = R_N = 2 Ω 

Thevenin voltage is calculated as:

VTH = I_N × R_N

VTH = 3 × 2 = 6 V

Concept:

Thevenin's Theorem:

Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

To find V_oc: Calculate the open-circuit voltage across load terminals. This open-circuit voltage is called Thevenin’s voltage (V_th).

To find I_sc: Short the load terminals and then calculate the current flowing through it. This current is called Norton current (or) short circuit current (i_sc).

To find R_th: Since there are Independent sources in the circuit, we can’t find Rth directly. We will calculate Rth using Voc and Isc and it is given by

\({{\rm{R}}_{{\rm{th}}}} = \frac{{{{\rm{V}}_{{\rm{oc}}}}}}{{{{\rm{i}}_{{\rm{sc}}}}}}\)  

Application:

Given: Load line equation = V + i = 100

To obtain open-circuit voltage (Vth) put i = 0 in load line equation 

⇒ Vth = 100 V

To obtain short-circuit current (isc) put V = 0 in load line equation

⇒ isc = 100 A

So, \({R_{th}} = \frac{{{V_{th}}}}{{{i_{sc}}}} = \frac{{100}}{{100}} = 1{\rm{\Omega }}\)

Equivalent circuit is

Current (i) = 100/2 = 50 A

Applying loop-law in the given circuit.

- V + i × R = 0

- V + I × 1 = 0

⇒ V = i

Given Load line equation is V + i = 100

Putting V = i 

then i + i = 100 

⇒ i = 50 A

Concept:

According to Thevenin’s theorem, any linear circuit across a load can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth as shown:

Vth = Open circuit Voltage at a – b (by removing the load), i.e.

R_th = equivalent resistance obtaining by deactivating sources.(Independent voltage source is short-circuited

and Independent current source is open  circuited)

Calculation:

Calculation of V_th

Apply KVL in the loop

20 - 3I - 6I + 10 = 0

I = (20 + 10) / (3 + 6)

I = 30 / 9

V_6Ω  = (30 / 9) × 6 

V6Ω = 20 V

Now apply KVL in the outer loop

- 10 + V_6Ω  - V_ab = 0

V_ab = V_th = (20 - 10) V = 10 V

where, I = current through loop 

Calculation of R_th

R_th = [3 || 6] + 3

Rth = 2 + 3 = 5 Ω

Fill Factor (FF):

Fill factor (FF) is the ratio of the actual maximum obtainable power (Pmax) to the product of short circuit current (I_sc) and open circuit voltage (V_oc).

We have,

actual maximum obtainable power (Pmax ) = 30W

Open circuit voltage (Voc) = 10 V

short circuit current (Isc) = 5 A

From the above concept,

\(FF=\frac{P_{max}}{V_{oc}I_{sc}}\)

or, \(FF=\frac{30}{10\times 5}=\frac{3}{5}=0.6\)

Given the circuit diagram:

To find Thevenin’s resistance, there is a need to suppress all independent sources with their internal resistance

Now, the circuit becomes

Now, introduce a current source of 1A between terminals A and B

Apply nodal analysis at node V_AB

\(\frac{{{V_{AB}} - \left( { - 3{V_{AB}}} \right)}}{{1K{\rm{Ω }}}} + \frac{{{V_{AB}}}}{{1k{\rm{Ω }}}} - 1 = 0\)

⇒ 4 V_AB + V_AB = 1000

⇒ V_AB = 200 V

Now Thevenin's resistance 

R_th = (V_AB / 1 A) = 200 Ω = 0.2 kΩ 

Therefore, the Thevenin’s resistance across terminals A and B is 0.2 kΩ 

Thevenin's Theorem:

Thevenin’s Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance connected across the load".

Thevenin's equivalent circuit:

Calculation of R_th:

6 Ω and 4 Ω are in parallel

\(R_{th} = {6\times 4 \over 6+4}\)

R_th = 2.4 Ω 

Calculation of Vth:

\( {V_{th} -10\over 6}+ {V_{th} -5\over 4}=0\)

\({2V_{th}-20+3V_{th}-15\over 12}=0\)

5V_th = 35

V_th = 7 V

Concept:

According to Thevenin’s theorem, any linear circuit across a load can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth as shown:

Vth = Open circuit Voltage at a – b (by removing the load), i.e.

If a linear circuit contains dependent sources only, i.e., there is no independent source present in the network, then the open-circuit voltage or Thevenin voltage will simply be zero. (Since there is no excitation present)

Referred value in Transformer:

In order to simplify the calculation, it is theoretically possible to transfer the voltage, current, and impedance of one winding to the other winding and combined them to a single value for each quantity.

Considered a transformer has turns ration 'a' which is given by,

\(a = \frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_1}}}{{{I_2}}}\)

Where,

I1 and I2 are primary and secondary current respectively.

V1 and V2 are primary and secondary voltage respectively.

N1 and N2 are numbers of turn in primary and secondary respectively.

For an ideal transformer:

Input Power = Output Power

\(\frac{{V_1^2}}{{{Z_1}}} = \frac{{V_2^2}}{{{Z_2}}}\)

\(\frac{{V_2^2}}{{V_1^2}} = \frac{{{Z_2}}}{{{Z_1}}}\)

\({\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^2} = \frac{{{Z_2}}}{{{Z_1}}}\)

\({a^2} = \frac{{{Z_2}}}{{{Z_1}}}\)

Equivalent secondary Impedance in terms of primary Impedance:

\({Z_1} = \frac{{{Z_2}}}{{{a^2}}}\)

Equivalent primary Impedance in terms of secondary Impedance:

\({Z_2} = {a^2}{Z_1}\)

Calculation:

In the given circuit diagram Thevenin's resistance referred from the x-y side is nothing but the impedance of the transformer from the secondary side

​

Thevenin's resistance, 

Z_th = a2 Z1 = (2)² × 1Ω 

Zth = 4 Ω 

Thevenin's voltage is nothing but voltage from the secondary side 

V_xy = V_th

\(\frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}} \)

V₂ = V₁ × (N₂ / N₁)

⇒ V₂ = V_th = sin ωt × (2 / 1) = 2 sin ωt

Therefore the Thevenin's voltage equivalent voltage and impedance as seen from the terminals x and y for the circuit is 2 sin (ωt), 4 Ω

Concept:

Thevenin’s voltage (VOC) is the open-circuit voltage across the load terminals.

Thevenin’s resistance (Rth) is the equivalent resistance across load terminals by replacing all the independent voltage sources with short circuit and all the independent current sources with an open circuit.

Calculation:

Thevenin voltage:

By applying the KVL,

-50 + 10I + 15I + 100 = 0

⇒ I = -2 A

V_TH = V_AB = 50 – 10I = 70 V

Thevenin resistance:

R_AB = R_TH = (10||15) = 6Ω

Concept:

Thevenin's Theorem:

• In a linear circuit containing several voltages(EMFs) and resistances.
• It can be replaced by just one single voltage(V_s) in series with a single resistance(R_s) connected across the load.
• Where V_s is Thevenin's Voltage also represented as V_th
• R_s is Thevenin's resistance.

For calculation of Thevenin's Voltage(V_th):

• Remove the load resistance.
• Find the open-circuit voltage across the load resistance terminals.

Calculation:

After removing the load resistance across terminals X-Y, the circuit will be:

V_th will be the voltage across terminals X-Y

By taking KVL in the outer loop of the circuit as shown:

- 12 + 6 (I_x) + 3 (4I_x) = 0

18 I_x = 12

I_x = 2/3 A

The Thevenin's voltage is the voltage across 3 Ω resistance

V_th = 3 (4I_x)

= 12 I_x (I_x = 2/3 A)

= 12 × 2/3

= 8 V

The value of Vth across X-Y terminals is 8 V

Concept:

Thevenin's Theorem:

Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

To solve Problems with independent Sources:

To find Voc: Calculate the open-circuit voltage across load terminals. This open-circuit voltage is called Thevenin’s voltage (Vth).

To find Rth: 

To calculate Thevenin’s Resistance we should replace all independent current sources with an Open circuit and Independent voltage sources with a Short circuit.

Application:

Let 2 Ω be the Load of the given circuit and open circuit to get Thevenin voltage.

A circuit diagram can be drawn as

By applying KVL,

9 - 4I - 4I - 6I = 0

\(I=\dfrac{9}{4+4+6}=\dfrac{9}{14}\ A\)

\(V_{th}=4I=4\times \dfrac{9}{14} =2.57\ V\)

Concept:

Thevenin's Theorem:

Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

To solve Problems with Dependent Sources:

To find Voc : Calculate the open-circuit voltage across load terminals. This open-circuit voltage is called Thevenin’s voltage (Vth).

To find Rth: 

To calculate Thevenin’s Resistance we should replace all independent current sources by Open circuit and Independent voltage sources by Short circuit (keep dependent sources as it is).

Circuits that have only dependent sources can’t function on their own so Vth and Isc don’t exist but still they exhibit resistance, that resistance can be indirectly determined by V/I method by placing an active source across the terminals.

1. Place a voltage source of 1V across the terminal and find the current (IT) flowing through it. Then,  

\({{\rm{R}}_{{\rm{th}}}} = {{\rm{R}}_{\rm{N}}} = \frac{{1{\rm{V}}}}{{{{\rm{I}}_{\rm{T}}}}}\)

(or) 

2. Place a current source of 1A across the terminals and find the voltage (Vt) across the current source. Then,

 \({{\rm{R}}_{{\rm{th}}}} = {{\rm{R}}_{\rm{N}}} = \frac{{{{\rm{V}}_{\rm{T}}}}}{{1{\rm{A}}}}\)

Calculation:

Thevenin voltage (V_th):

V_th is the voltage across terminal x-y

By applying nodal analysis at node A,

\(\frac{{{V_{th}} - 2i}}{1} + \frac{{{V_{th}}}}{1} + \frac{{{V_{th}}}}{2} = 2\)

\(2{V_{th}} - 2i + \frac{{{V_{th}}}}{2} = 2\)

And V_th = i

⇒ V_th = 4V

Thevenin resistance (R_th):

To find the Thevenin’s resistance, we need to short circuit the ideal voltage source and open circuit the independent (ideal) current source.

As the given circuit contains a dependent source, then place a current source of 1A across the terminals and find the voltage (Vx) across the current source. Then

By applying the nodal analysis, at node A

\(\frac{{{V_x} - 2i}}{1} + \frac{{{V_x}}}{1} + \frac{{{V_x}}}{2} = 1\)

\( \Rightarrow 2{V_x} - 2i + \frac{{{V_x}}}{2} = 1\)

And V_x = 1

⇒ V_x = 2V

\({R_{th}} = \frac{{{V_x}}}{1} = \frac{2}{1} = 2\;{\rm{\Omega }}\)

Thevenin equivalent circuit for the given circuit is,