KCL MCQ Quiz - Objective Question with Answer for KCL - 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

Concept

The value of current across any branch is given by:

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

where, I = Current

V = Voltage

R = Resistance

Calculation

Given, V = 10 volts

R₁ = 1 Ω, R₂ = 2Ω and R₃ = 5Ω 

In parallel, the voltage remains the same.

\(I_1={10\over 1}=10A\)

\(I_2={10\over 2}=5A\)

\(I_3={10\over 5}=2A\)

Kirchhoff's Current Law

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

KCL is based on the conservation of charge.

\(-{dq_1\over dt}-{dq_2\over dt}-{dq_3\over dt}+{dq_4\over dt}+{dq_5\over dt}+{dq_6\over dt}=0\)

Hence, statement 4 is correct.

Kirchhoff's Voltage Law

According to Kirchhoff's Voltage Law, the sum of the total voltage drop in a closed loop is equal to zero.

\(V=V_1+V_2.....V_n\)

\(V=I(R_1+R_2...R_n)\)

Kirchhoff's Current Law

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

KCL is based on the conservation of charge.

Explanation

In any electric circuit, if i₁ and i₃ are incoming currents and i₂ and i₄ are outgoing currents from a node, then:

Sum of incoming current = Sum of outgoing current

i₁ + i₃ = i₂ + i₄

Additional Information 
Kirchhoff's Voltage Law

According to Kirchhoff's Voltage Law, the sum of the total voltage drop in a closed loop is equal to zero.

\(V=V_1+V_2.....V_n\)

\(V=I(R_1+R_2...R_n)\)

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:

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

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

Concept:

Kirchoff's Current Law (KCL): The algebraic sum of the currents at a node is zero. Alternatively, the sum of the currents entering a node is equal to the sum of the currents leaving that node.

I₁ - I₂ - I₃ + I₄ + I₅ = 0

Calculations:

Circuit Diagram:

Here, i_a is given by,

\({i_a} = \frac{{{V_x} - 12}}{4}\)     ...(1)

Apply KVL at node V_x,

\(\frac{{{V_x} - 12}}{4} + 3{i_a} + \frac{{{V_x}}}{2} = 0\)

Put the value of i_a in above equation,

\(\frac{{{V_x} - 12}}{4} + 3\left( {\frac{{{V_x} - 12}}{4}} \right) + \frac{{{V_x}}}{2} = 0\)

\(\frac{{{V_x} - 12}}{4} + \frac{{3{V_x} - 36}}{4} + \frac{{{V_x}}}{2} = 0\)

V_x  - 12 + 3V_x - 36 + 2V_x = 0

6 V_x = 48

V_x = 8 V

Put this value of V in equation (1)

i_a = (8 - 12) / 4 = -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.

Considered the node in the figure shown below, in which i1, i3, and i5 are incoming current and i2, i4, and i6 are outgoing current.

Applying KCL gives,

i1 + i3 + i5 = i2 + i4 + i6

Hence, the sum of currents entering a node is equal to the sum of the currents leaving the node.

Note: KCL is based on the conservation of charge.

Application:

Applying KCL to the circuit,

I_T + I₂ = I₁ + I₃

Given,

I1 = 10 A, I2 = 15 A and I3 = 20 A

Hence,

I_T = 10 + 20 - 15 = 15 A

CONCEPT:

• Kirchhoff’s Current Law:  The total current entering a junction or a node is equal to the charge leaving the node as no charge is lost.
• The algebraic sum of every current entering and leaving the node has to be null. This property of Kirchhoff law is commonly known as Conservation of charge.
• Node: Node is a point in a network where two or more circuit elements are connected.

I1 + i2 + i3 + i4 + i5 = 0

EXPLANATION:

Calculation:

By applying KCL, at node a

i1 – i2 – i3 + i4 = 0

Additional Information

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

By applying voltage division in the input loop,

\({V_x} = 18\left( {\frac{6}{{12 + 6}}} \right) = 6\;V\)

i₀ will be zero, as there is no closed path.

By applying KCL in the right-side circuit,

\(\frac{{{V_x}}}{2} + {i_1} + {i_2} = 0\)

\(\Rightarrow {i_1} + {i_2} = - \frac{{{V_x}}}{2} = - 3\) ----(1)

As 10 Ω and 5 Ω are connected in parallel, voltage across them are equal.

⇒ 10 i₁ = 5 i₂ ⇒ i₂ = 2i₁ ----(2)

From equation (1) and (2)

Calculation:

When source voltage is applied then

Applying source transformation in the current source

Applying Nodal Analysis

\(\rm \frac{V-1}{2}+\frac{V}{2}+\frac{V-2}{1}=0\)

or, \(\rm \frac{V-1+V+2V-4}{2}=0\)

or, 4 V = 5 ⇒ V = 1.25 V

∴ \(\rm I=\frac{1.25}{2}=0.625\ A\)

Therefore, the correct option is (4).