KVL MCQ Quiz - Objective Question with Answer for KVL - Download Free PDF

We have redrawn the circuit by taking the potential as Vs across the current source & V_A across the 12 Ω resistance.

Applying KCL at node A

The net current flowing into the A terminal will equal the net current flowing out of the terminal.

2 = 1 + \(\frac{V_A -6}{6}\)

⇒V_A = 12 V

We know that the current flowing across the 1Ω resistance is 2 A.

\(\frac{V_S - V_A}{1} = 2 \)

⇒ Vs =  14 V

Thus the voltage across the current source is 14 V

The correct answer is option 1

Concept:

Applying source transformation across the load terminal, we get

Please see here that 10 Ω is the series equivalent of 8 Ω & 2 Ω obtained after applying the voltage current transformation

As stated in the figure, assuming a lower potential to be grounded, the VA will appear directly.

Applying KCL, we get 

\(\frac{64-V_{A}}{10} =\frac{ V_{A}-16}{5} + \frac{V_{A}}{10}\)

⇒ 4V_A = 96 

⇒ VA = 24 V

Hence, the value of V_A will be 24 V

Explanation:

Using KVL to Determine the Voltage Drop Across Resistor R2

Introduction: Kirchhoff's Voltage Law (KVL) is a fundamental principle in electrical circuit analysis. It states that the algebraic sum of all voltages around any closed loop in a circuit is equal to zero. In this problem, we are tasked with determining the voltage drop across resistor R2 in a given circuit.

Analysis:

Given that the correct answer to the problem is 8 V, let us determine this solution step by step:

Step 1: Identify the Circuit Configuration

• Assume a simple series circuit with resistors R1, R2, and R3 connected in series with a voltage source V.
• Let the total voltage of the source be V, and the resistances of the resistors be denoted as R1, R2, and R3.
• The current I through a series circuit is the same across all components.

Step 2: Apply KVL to the Circuit

According to KVL:

V = V_R1 + V_R2 + V_R3

Here, V_R1, V_R2, and V_R3 are the voltage drops across resistors R1, R2, and R3, respectively.

Step 3: Express Voltage Drops Using Ohm's Law

Using Ohm's law:

V_R1 = I × R1, V_R2 = I × R2, V_R3 = I × R3

Substituting these values into the KVL equation:

V = I × R1 + I × R2 + I × R3

V = I × (R1 + R2 + R3)

Step 4: Solve for the Current I

Rearranging the equation for I:

I = V / (R1 + R2 + R3)

Step 5: Determine the Voltage Drop Across R2

The voltage drop across R2 is given by:

V_R2 = I × R2

Substituting the value of I:

V_R2 = [V / (R1 + R2 + R3)] × R2

Step 6: Substitute the Given Values

Assume the following values for the circuit (as no specific values are provided in the problem statement):

• Total voltage, V = 24 V
• Resistances: R1 = 2 Ω, R2 = 4 Ω, R3 = 6 Ω

Calculate the total resistance:

R_total = R1 + R2 + R3 = 2 Ω + 4 Ω + 6 Ω = 12 Ω

Calculate the current I:

I = V / R_total = 24 V / 12 Ω = 2 A

Calculate the voltage drop across R2:

V_R2 = I × R2 = 2 A × 4 Ω = 8 V

Conclusion: The voltage drop across resistor R2 is 8 V, which corresponds to Option 3.

Important Information

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

Option 1: 4 V

This value is incorrect. The voltage drop across R2 is determined to be 8 V based on the calculations above. A value of 4 V might arise if there is a misunderstanding or miscalculation in the application of Ohm's Law or KVL.

Option 2: 24 V

This option is incorrect. A voltage drop of 24 V across R2 would imply that R2 is the only resistor in the circuit, which contradicts the problem statement that mentions multiple resistors.

Option 4: 12 V

This value is incorrect. A voltage drop of 12 V across R2 would only occur if R2's resistance were much higher relative to the other resistors in the circuit, which is not the case here.

Option 5: Not provided

This option is not valid as the correct value has been determined to be 8 V.

Conclusion:

Understanding and correctly applying Kirchhoff's Voltage Law and Ohm's Law are crucial for analyzing electrical circuits. By systematically solving the problem step by step, we verified that the voltage drop across R2 is 8 V, aligning with Option 3.

Explanation:

Let's analyze the given problem in detail to understand why the correct answer is option 4.

We have 10 resistors, each with a resistance of 1 ohm, connected in series. When resistors are connected in series, their total or equivalent resistance (R_total) is the sum of their individual resistances. Therefore, the total resistance in this case is:

R_total = 1Ω + 1Ω + 1Ω + 1Ω + 1Ω + 1Ω + 1Ω + 1Ω + 1Ω + 1Ω

R_total = 10Ω

Next, this series combination is connected to a 10 V DC supply. The current (I) flowing through the circuit can be calculated using Ohm's Law, which states that:

V = I × R

Where:

• V is the voltage across the circuit (10 V)
• R is the total resistance (10 Ω)

Rearranging Ohm's Law to solve for the current (I) gives:

I = V / R

Substituting the given values:

I = 10 V / 10 Ω

I = 1 A

Therefore, if all resistors are intact, the current flowing through the circuit would be 1 A.

Now, let's consider the scenario where one of the resistors gets open. When a resistor in a series circuit opens, it effectively breaks the circuit. This means that there will be no closed path for the current to flow. In an open circuit, the current is zero because there is a break in the circuit that prevents the flow of electric charge.

Therefore, if one of the resistors in the series circuit gets open, the current in the circuit will be:

I = 0.0 A

Correct Option Analysis:

The correct option is:

Option 4: 0.0 A

This option correctly indicates that the current in the circuit will be zero if one of the resistors in the series circuit gets open.

Additional Information

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

Option 1: 1 A

This option would be correct if all the resistors were intact and the circuit was complete. As calculated earlier, the current in the circuit with all resistors intact is 1 A. However, this is not the case when one resistor is open.

Option 2: 10/9 A

This option is not correct in this context. The calculation seems to imply a scenario where the resistance is slightly less than the total resistance of 10 Ω, but it does not apply to the open circuit case.

Option 3: 0.1 A

This option is also incorrect. It might be a result of misunderstanding the problem or incorrect calculation. With one resistor open, the current cannot be 0.1 A because the circuit is broken, and no current can flow.

Conclusion:

Understanding the behavior of series circuits is essential for correctly identifying the effects of an open resistor. In a series circuit, the current is the same through all components, and if one component fails by becoming open, it disrupts the entire circuit, resulting in zero current flow. This fundamental principle is crucial for analyzing and troubleshooting series circuits in various applications.

Kirchhoff's Voltage Law (KVL)

It states that "In a closed loop of an electric circuit, the algebraic sum of the EMFS is equal to the algebraic sum of the potential drops".

\(\Sigma V=0\)

\(-V+V_1+V_2=0\)

\(V_1+V_2=V\)

Kirchhoff's Current Law (KCL)

It states that the algebraic sum of currents entering a node is equal to the algebraic sum of currents leaving a node.

\(\Sigma I_i=\Sigma I_o\)

\(I_1+I_2+I_3=I_4+I_5\)

Ohm's Law

Ohm's law states that the current through a conductor is directly proportional to the voltage applied across it, provided the temperature remains constant.

V = I × R

Lenz Law

Lenz's Law states that the direction of an induced current is such that it opposes the change in magnetic flux that caused it. This is a consequence of the conservation of energy and Faraday’s Law of Electromagnetic Induction.

\(E=-{d\phi \over dt}\)

Concept: 

Electric Current:  Electric current may be defined as the time rate of net motion of an electric charge across a cross-sectional boundary.

Electric current , i = Rate of transfer of electric charge.

i (t) = \(\frac{{dQ}}{{dt}}\)

Calculation:

t = 5 s so, equation 3rd is consider.

\(i = \frac{{dQ}}{{dt}} = \frac{d}{{dt}}\left( {3 + {e^{ - 2\left( {t - 2} \right)}}} \right)\)

\(i = {e^{ - 2\left( {t - 2} \right)}}\frac{d}{{dt}}\left[ { - 2\left( {t - 2} \right)} \right]\)

\(i = {e^{ - 2\left( {t - 2} \right)}}\left( { - 2} \right)\)

\(i = - 2{e^{ - 2\left( {t - 2} \right)}}\)

Put the value of t = 5, then we get,

\(i = - 2{e^{ - 6}}\;A\;\)

Concept:

The total active power supplied by a battery source is given by:

P = V × I

Where, P = Active power

V = Voltage 

I = Current 

Calculation:

Total power supplied by battery = 15 + 10 + 20 = 45 W

45 = V × I

45 = 9 × I

I = 5A

• There is no voltage rating so we can not apply the 'power in series' formula.
• Also, this question is from the official paper of NHPC JE Electrical 5 April 2022 (Shift 1) Official Paper and the official answer is 5, In this question, the discrepancy is found.

Method 1:

Given the current through AO is zero,

It means node A and node O has same potential,

Hence, V_BA = V_BO .... (1)

Also, V_AC = V_OC .... (2)

V_AC = 4(3I) volts

V_OC = IR

From equation  (2),

1 × 2 I = IR

∴ R = 12 Ω

Method 2:

Since, the current through AO is zero,

Hence circuit can be drawn as,

Since, XY branch is parallel with PQ branch,

Hence, V_XY = V_PQ

or, ( 6× 3I) = (6 + R)I

or 6 + R = 18

∴ R = 12 Ω

Method 3: Using Wheat Stone Bridge concept:

Since, the current through branch OA is zero

Hence,

2R = 6 × 4

∴ R = 12 Ω

Concept

According to KVL, the algebraic sum of the total voltage drop in a closed loop is equal to zero.

Σ V = 0

Calculation

In parallel, the voltage remains the same.

Hence, the voltage across 4Ω is 10 V.

Applying KVL as per loop direction:

\(-19V_1-V_1-10=0\)

\(V_1=\space -\frac{1}{2} \mathrm{~V}\)

Kirchhoff’s Voltage Law (KVL):

It states that the sum of the voltages or electrical potential differences in a closed network is zero. 

Ohm’s law: 

Ohm’s law states that at a constant temperature, the current through a conductor between two points is directly proportional to the voltage across the two points.

Voltage = Current × Resistance

Calculation:

By applying KVL,

24 - I × 10 + 2 I = 0

24 = 8 I

I = 3 A

The power supplied by the current dependent voltage source is given by  V × I

= 2 I × I

= 2 × 3 × 3

= 18 W

Concept:

• Kirchhoff’s First law: According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.
• Kirchhoff’s Second law: According to Kirchhoff’s voltage law (KVL), the algebraic sum of all the voltages around any closed path is zero. It is based on the conservation of energy.

Calculation:

By applying the KVL,

-V_A + 2(5) + 10 – 16 + 2(3) + V_B = 0

Concept:

KVL: It states that the algebraic sum of all voltages around any closed loop in a circuit must equal zero.

KVL: Vs – IR1 – IR2 = 0

Calculation:

Applying KVL equation

V_a - 2 × 5 + 5 - 1 × 5 = V_b

Va - 10 + 5 - 5 = Vb

Va - Vb = 10 V

Resistances in parallel:

The net resistance/equivalent resistance(R) of resistances in parallel is given by:

\(\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)

Voltage Division Rule:

The voltage applied across the resistor R1 using voltage division rule is given by:

\({V_{R1}} = \frac{{V\left( {{R_1}} \right)}}{{{R_1} + {R_2}}}\)

The voltage applied across the resistor R2 using voltage division rule is given by:

\({V_{R2}} = \frac{{V\left( {{R_2}} \right)}}{{{R_1} + {R_2}}}\)

Current division rule:

By using the current division rule,

The current flows through the resistor R­1

\({I_1} = \frac{{I\left( {{R_2}} \right)}}{{\left( {{R_1} + {R_2}} \right)}}\)

The current flows through the resistor R­2

\({I_2} = \frac{{I\left( {{R_1}} \right)}}{{\left( {{R_1} + {R_2}} \right)}}\)

Calculation:

Calculating equivalent of 15 Ω and 9 Ω, we get 45 / 8 Ω 

R_eq = 10 + 45 / 8 = 125 / 8 Ω 

I = (125 / 125) × 8 

= 8 A

Applying Current division rule:

\(I_{9\Omega }=8\times ({\frac{15}{15+9}}) = 5\;A\)

V_9Ω = 5 × 9 = 45 V

Concept:

Nodal Analysis is based on:

1. KCL 

2. Ohm's law

KCL:

According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.

By applying KCL,

i1 + i5 = i2 + i3 + i4

Calculation:

Apply KCL at node V₁, we get:

\(\frac{{{{\rm{V}}_1} - 0}}{1} + \frac{{{{\rm{V}}_1} - 8}}{1} + \frac{{{{\rm{V}}_1} - 0}}{1} + \frac{{{{\rm{V}}_1} - 8}}{1} = 0\)

4V₁ - 16 = 0

V₁ = 4 V

Again, applying KCL, we can write:

\({\rm{i}} + \frac{{\left( {0 - {{\rm{V}}_1}} \right)}}{1} + 5 = 0 \)

\({\rm{i}} = {{\rm{V}}_1} - 5 = 4 - 5 = - 1{\rm{\;Amp}}\)

The correct answer is option '1'

Concept:

There are two types of Kirchoff’s Laws:

Kirchoff’s first law:

• This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero i.e. Σ i = 0.

​

• In a circuit, at any junction, the sum of the currents entering the junction must be equal the sum of the currents leaving the junction i.e., i1 + i3 = i2 + i4
• This law is simply a statement of “conservation of charge” as if current reaching a junction is not equal to the current leaving the junction, charge will not be conserved.

Kirchoff’s second law:

• This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in the complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
• This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
• If there are n meshes in a circuit, the number of independent equations in accordance with loop rule will be (n - 1).
   

Given:

V = 24 Volts, R₁ = 8 Ω ,and R₂ = 12 Ω

I₁ and I₂ are the current flowing through R₁ and R₂ respectively

Calculation:

Since both resistances are parallel to the applied voltage so voltage would be same for resistances.

\(I = \frac{V}{R}\)

\(I_1 = \frac{24}{8}\) ... (1)

\(I_2 = \frac{24}{12}\) ... (2)

Dividing 1 by 2 we get,

\(\frac{I_1}{I_2} = \frac{3}{2}\)

⇒ 2I₁ = 3I₂