Kirchhoff's Laws MCQ Quiz - Objective Question with Answer for Kirchhoff's Laws - Download Free PDF

Explanation:

Analysis of the Circuit to Determine the Value of Current (I):

In the given problem, we are required to determine the value of the current (I) in the circuit. However, based on the provided options and the correct answer being Option 2 (Indeterminate), it suggests that the circuit configuration or the data provided in the problem is insufficient to calculate the current. Let us analyze why this is the case and why the other options are incorrect.

Reason for Option 2 (Indeterminate) Being Correct:

To determine the current (I) in a circuit, certain essential information is required, such as:

• The complete circuit diagram, which includes the arrangement of components like resistors, voltage sources, and their respective values.
• Ohm's Law or Kirchhoff's Laws for analyzing the circuit.
• Boundary conditions or constraints, if any.

However, in this case, the problem does not provide sufficient details about the circuit configuration, such as the values of resistances, voltage sources, or the overall circuit layout. Without this critical information, it is impossible to apply the necessary electrical laws to calculate the current. Hence, the value of current (I) cannot be determined from the given data, making the correct answer Option 2: Indeterminate.

Important Information:

Let us analyze why the other options are incorrect:

Option 1: 2 A

This option suggests that the current in the circuit is 2 A. However, without knowing the specific values of resistances, voltage sources, or the circuit configuration, there is no basis to claim that the current is precisely 2 A. This is purely speculative and cannot be justified with the given information.

Option 3: -1 A

This option assumes that the current has a value of -1 A. The negative sign might indicate the direction of current flow, but again, without any data about the circuit elements or their arrangement, there is no justification for this value. Thus, this option is also incorrect.

Option 4: 1 A

This option states that the current in the circuit is 1 A. Like Option 1, this is an arbitrary value that cannot be verified in the absence of information about the circuit's resistances, voltage sources, or layout. Therefore, this option is not valid.

Option 5: (Not provided in the problem statement)

The problem does not explicitly provide details for Option 5. However, any assumption about the value of the current without proper data is inherently flawed and cannot be considered correct.

Conclusion:

The inability to determine the current (I) in the circuit arises from a lack of critical information regarding the circuit's components and their configuration. In such cases, it is crucial to recognize that the problem is indeterminate, as no definitive calculation can be performed. This highlights the importance of providing complete circuit details when solving electrical circuit problems.

Kirchhoff's Current Law

According to Kirchhoff's Current Law, the total current entering a junction or a node is equal to the current leaving the node. The algebraic sum of every current entering and leaving the node has to be zero.

Calculation

Assuming the sign of the incoming current as negative and the outgoing current as positive.

-2 - 20 + 5 + I₄ = 0 

I₄ = 17 A

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.

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

Kirchoff's Current Law (KCL)

According to KCL, the algebraic current at a node is always zero.

ΣI = 0

\(-I_1-I_2-I_3+I_4+I_5=0\)

\(-{dq_1\over dt}-{dq_2\over dt}-{dq_3\over dt}+{dq_4\over dt}+{dq_5\over dt}=0\)

Hence, KCL is based on the conservation of charge principle.

Kirchoff's Voltage Law (KVL)

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

ΣV = 0

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

\(\int E.dl=0\)

A conservative field means that the work done by the electric field on a charge depends only on the starting and ending points, not on the path taken.

For a closed loop, the starting and ending points are the same, so the total work done (or total potential difference) is zero.

Hence, KCL is based on the conservation of electric field.

Concept:

Kirchhoff’s Current Law (KCL)

It states that the amount of current flowing into a node or junction is equal to the sum of the currents flowing out of it

Current entering through a load has a positive polarity and exit point has a negative polarity and for source, it will be reversed.

Calculation:

Let current through 3 Ω is I₁ and through 2 Ω is I₂ and current through 5 V is I ampere respectively as shown below

\(I_1 = \frac{{6~V}}{{3{\rm{~\Omega }}}} = 2~A\)

\(I_2 = \frac{{2V}}{{2{\rm{Ω }}}} = 1A\)

Apply KCL at N/W N₂,

I + I2 = I1

I + 1 = 2

I = 1 A

P = 5 × 1 = 5 W

Kirchhoff's current law (KCL) is applicable to networks that are:

• Kirchhoff’s law is applicable to both AC and DC circuits. It is not applicable for time-varying magnetic fields.
• Unilateral or bilateral 
• Active or passive 
• Linear or non-linear
• Lumped network

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:

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.

For any load, current enter through positive polarity and exit from negative polarity and for source it will be reversed.

Calculation:

Apply KVL in the loop as shown below,

The current flow through 2 Ω is 5 A as shown in fig and it will give a drop of 10 V.

From the figure we get,

- V_out + 10 + 10 = 0

Vout = 20 V

Concept:

KCL in DC circuits:

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.

KCL in AC circuits:

The Kirchhoff’s current law as applied to the ac circuit is defined as the phasor sum of currents entering the node is equal to the phasor sum of currents leaving the node.

Calculation:

KCL at Node A

I₁ + 1A = 3A

I₁ = 2 A

KCL at node B

I₂ + 2A = 6A

I₂ = 4 A

KCL at node C

I₃ = 4A + 2A

I₃ = 6A

KCL at node D

I = 6A - 5A = 1 A

Kirchhoff’s Current Law (KCL): It states that the algebraic sum of currents entering a node or closed boundary is zero.

Mathematically, KCL implies that

\(\mathop \sum \limits_{n = 1}^N {i_n} = 0 \)

Where N is the number of branches connected to the node

And in is the nth current entering or leaving the node.

By this law current entering a node may be regarded as positive and current leaving a node may be regarded as negative or vice-versa.

Note: KCL is based on the conservation of charge.

Calculation:

Given circuit,

Apply KCL,

2 + 4 = 10 + I_o

or, 6 = 10 + I_o

∴ I_o = 6 - 10 = - 4 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.

Concept:

Kirchhoff’s Current Law:

It says the total current entering a circuits junction is exactly equal to the total current leaving the same junction.

Or sum of all current entering or leaving a node is zero

According to KCL: ∑I = 0

Applying KCL:

I1 + I2 – I3 – I4 – I5 = 0

I1 + I2 = I3 + I4 + I5

Calculation:

Apply KCL at node 'a'

i₀ = 0.5i₀ + 3

0.5i₀ = 3

i₀ = 6 A

∴v₀ = 4i₀ = 4× 6 = 24 V

Let voltage across the 8 Ω resistance is 'V' volt.

∴ Current across the 8 Ω is given by

\(I = {{V} \over 8}\)

Now by applying KCL at node V, we get

\({{V - 5} \over 2}+{{V +3} \over 4}+{{V } \over 8}=0\)

4V - 20 + 2V + 6 + V = 0

\(V = {{14} \over 7} =2~volt\)

Now current flowing through the 8 Ω resistance is

\(I = {{2} \over 8}\)

I = 0.25 Amp

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

Node a is directly connected to the ground, so the voltage at node a is equal to 0 V.

V_a = 0 V

Applying the KCL at node a, we get:

\({0-6\over 10}+i_a+5i_a=0\)

i_a = 0.1 A

Also, V_a - V_b = 20 × (5i_a)

0 - Vb = 20 × 5 × 0.1

Vb = −10 V