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Kirchhoffs Rules

Learn Kirchhoff’s rules and Kirchhoff’s voltage, first, second, circuit law. Practice equation to calculation of current using Kirchhoff's voltage, first, second, circuit law.

Kirchhoff's First Law

  • At any junction the sum of currents (entering or leaving) must be zero.

      \(\displaystyle\sum _{junction} I=0\)

  • Kirchhoff's first  law is based upon the conservation of electric charge.
  • All charges that enter a given point in a circuit must leave that point because charge cannot build up at a point.

                 \(\sum I _{Enter} = \sum I _{Leave}\) 

                  \(I_{1} = I_{2}+I_{3}\)

Illustration Questions

Two resistors \(R_1 = 2\,\Omega\) and \(R_{2} = 3\,\Omega\) are connected as shown in figure. The total current  \(I = 5\,A\) is entering at point P. Calculate the current \((I_{2}) \)in resistor R2  if current in resister R1 is  \(I_{1} = 3\,A\).

A 7 A

B 2 A

C 3 A

D 8 A

×

By Kirchhoff's first law, total current across a junction is zero, i.e., amount of current entering a junction is same as amount of current leaving a junction.

 

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junction at point P

\(\sum I _{Enter} = \sum I _{Leave}\)

 

image

\(\sum I _{Enter} = I = 5\,A \)

\(\sum I _{Leave} = I_{1}+I_{2}= 3 + I_{2}\)

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\(I_{2} +3 = 5\)

\(\Rightarrow I_{2} = 5\,A - 3\,A\)

\(\Rightarrow I_{2} = 2\,A\)

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Two resistors \(R_1 = 2\,\Omega\) and \(R_{2} = 3\,\Omega\) are connected as shown in figure. The total current  \(I = 5\,A\) is entering at point P. Calculate the current \((I_{2}) \)in resistor R2  if current in resister R1 is  \(I_{1} = 3\,A\).

image
A

7 A

.

B

2 A

C

3 A

D

8 A

Option B is Correct

Calculation of Current using Kirchhoff's First Law

  • At any junction the sum of currents (entering or leaving) must be zero.

      \(\displaystyle\sum _{junction} I=0\)

  • Kirchhoff's first law is based upon the conservation of electric charge.
  • All charges that enter a given point in a circuit must leave that point because charge cannot build up at a point.

                 \(\sum I _{Enter} = \sum I _{Leave}\) 

                  \(I_{1} = I_{2}+I_{3}\)

Illustration Questions

Calculate \(I_{3}\) if \(I_{t} = 2\,A\), \(I_{1} = 0.75 \,A\) , \(I_{2} = 0.25\,A\) in  the given circuit.

A 1 A

B 2 A

C 3 A

D 1.5 A

×

 At Junction X

\(\sum I _{Enter} = \sum I _{Leave}\)

\(I_{t} = I_{1} + I_ {X} \)

 

 

image image

\(2\,A = 0.75\,A+ I_{X}\)

\(I_{X} = 2\,A - 0.75\,A\)

\(I_{X} = 1.25\,A\)

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At Junction Y

     \(\sum I_{ENTER} =\sum I_{LEAVE} \)

      \(I_{X} = I_{2}+I_{3}\)

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\(1.25 = 0.25 + I_{3}\)

     \(I_{3} = 1.25 -0.25 \)

     \(I_{3} = 1\,A\)

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Calculate \(I_{3}\) if \(I_{t} = 2\,A\), \(I_{1} = 0.75 \,A\) , \(I_{2} = 0.25\,A\) in  the given circuit.

image
A

1 A

.

B

2 A

C

3 A

D

1.5 A

Option A is Correct

Potential across Resistor

  • Voltmeter is an instrument used to measure the voltage across the resistor.
  • Voltmeter is always connected parallel to resistor.
  • Voltage across the resistor is the product of resistance of resistor and current through that resistor.

           \(\Delta V = IR\)

Illustration Questions

In the circuit, voltmeters are connected across each resistor. The voltmeter reading across R4 is not same as (consider current through voltmeter to be negligible)  

A Voltmeter reading across \( R_1\)

B \(R_4 I_6\)

C \(R_3 I_5\)

D None of these

×

Voltmeter reading across R1

\(\Delta V_1 = I_1\;R_1\)

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Voltmeter reading across R2

\(\Delta V_2 = I_2\;R_2\)

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Voltmeter reading across R3

\(\Delta V_3 = I_5\;R_3\)

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Voltmeter reading across R4

\(\Delta V_4 = I_6\;R_4\)

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Voltmeter reading across R3 and R4 are same as both are connected in parallel and in parallel, voltage is same.

\(\Delta V_3 = \Delta V_4\)

\(or,\,\, I_5\;R_3 = I_6\;R_4\)

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In the circuit, voltmeters are connected across each resistor. The voltmeter reading across R4 is not same as (consider current through voltmeter to be negligible)  

image
A

Voltmeter reading across \( R_1\)

.

B

\(R_4 I_6\)

C

\(R_3 I_5\)

D

None of these

Option A is Correct

Kirchhoff's Second Law

  • Kirchhoff's second law states that the sum of all potential differences across all elements around any closed circuit loop must be zero.

                       \(\displaystyle\sum_{\text{closed loop}} \Delta V =0\)

  • This law is based upon the principle of energy conservation.
  • A charge that moves around a closed path and returns to its starting point, has zero potential energy \((\Delta U=0)\).
  • Charge always moves from high potential end of resistor towards the low potential end.

Case 1 

If resistor is traversed along the direction of current, then potential difference \(\Delta V\) across the resistor is

                                          \(\Delta V= -IR\)

Direction of traversing = A to B

A= High potential

B= Low potential

Case 2

If resistor is traversed opposite to the direction of current, then potential difference \(\Delta V\) across resistor is

                         \(\Delta V = +IR\)

Direction of traversing = B to A

Case 3

If source of emf (assuming internal resistance = 0 ) is traversed in the direction of emf (–ve to +ve), then potential difference \(\Delta V\)  is

\(\Delta V = +\,\mathcal{E}\)

Case 4

If source of emf (assuming internal resistance =0) is traversed in the opposite direction of emf, then potential difference \(\Delta V\)  is

\(\Delta V = -\,\mathcal{E}\)

Illustration Questions

A single loop containing two resistors and two sources of emf is shown in figure. Choose the correct expression by Kirchhoff 's second law. 

A \(\mathcal{E} _1- IR_2 - \mathcal{E}_2 - IR_1 = 0\)

B \(-\mathcal{E} _1- IR_2 +\mathcal{E}_2 - IR_1 = 0\)

C \(\mathcal{E} _1+ IR_2 + \mathcal{E}_2 + IR_1 = 0\)

D \(-\mathcal{E} _1+ IR_2 - \mathcal{E}_2 + IR_1 = 0\)

×

For Resistor

Moving in the direction of current \(\Delta V = negative\) 

Moving opposite to direction of current  \(\Delta V = positive\)

image

For Battery 

Moving in the direction of emf (+ve to -ve)

\(\Delta V = -\mathcal{E}\)

Moving opposite to direction of emf (-ve to +ve)

\(\Delta V = +\mathcal{E}\)

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Traversing the circuit in clockwise direction, the sum of all potential differences across all elements around any closed loop must be zero.

\(\displaystyle\sum_{\text{closed loop}} \Delta V =0\)

image

\(\mathcal{E} _1- IR_2 - \mathcal{E}_2 - IR_1 = 0\)

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A single loop containing two resistors and two sources of emf is shown in figure. Choose the correct expression by Kirchhoff 's second law. 

image
A

\(\mathcal{E} _1- IR_2 - \mathcal{E}_2 - IR_1 = 0\)

.

B

\(-\mathcal{E} _1- IR_2 +\mathcal{E}_2 - IR_1 = 0\)

C

\(\mathcal{E} _1+ IR_2 + \mathcal{E}_2 + IR_1 = 0\)

D

\(-\mathcal{E} _1+ IR_2 - \mathcal{E}_2 + IR_1 = 0\)

Option A is Correct

Application of Kirchhoff's Second Law on a Circuit having Two Loops

  • Kirchhoff's second law states that the sum of all potential differences across all elements around any closed circuit loop must be zero.

                       \(\displaystyle\sum_{\text{closed loop}} \Delta V =0\)

  • This law is based upon the principle of  energy conservation.
  • A charge that moves around a closed path and returns to its starting point, has zero potential energy \((\Delta U=0)\).
  • Charge always moves from high potential end of resistor towards the low potential end.

Case 1

If resistor is traversed along the direction of current, then potential difference \(\Delta V\) across the resistor is

                                 \(\Delta V= -IR\)

Direction of traversing = A to B

A= High potential

B= Low potential

Case 2

If resistor is traversed opposite to the direction of current, then potential difference \(\Delta V\) across resistor is                                    \(\Delta V = +IR\)

          Direction of traversing = B to A

Case 3

If source of emf (assuming internal resistance =0 ) is traversed in the direction of emf (–ve to +ve), then potential difference \(\Delta V\)  is

\(\Delta V = +\mathcal{E}\)

Case 4

If source of emf (assuming internal resistance =0) is traversed in the opposite direction of emf, then potential difference \(\Delta V\)  is

\(\Delta V = -\mathcal{E}\)

Illustration Questions

Which option is correct by Kirchhoff's second law for the given circuit? \(\mathcal{E}_1>\mathcal{E}_2\)

A For loop \(I\) \(\mathcal{E} _1- I_1R_2 + \mathcal{E}_2 + I_1R_1 = 0\) For loop \(II\) \(\{-(I - I_1) R_3 \} + I_1R_2 = 0\)

B For loop \(I\) \(\mathcal{E} _1+ IR_2 + \mathcal{E}_2 + I_1R_1 = 0\) For loop \(II\) \(\{-(I - I_1) R_3 \} + I_1R_2 = 0\)

C For loop \(I\) \(\mathcal{E} _1- I_1R_2 - \mathcal{E}_2 - IR_1 = 0\) For loop \(II\) \(\{-(I - I_1) R_3 \} + I_1R_2 = 0\)

D For loop \(I\) \(\mathcal{E} _1- I_1R_2 + \mathcal{E}_2 + I_1R_1 = 0\) For loop \(II\) \(\{-(I - I_1) R_3 \} - I_1R_2 = 0\)

×

For Resistor

Moving in the direction of current \(\Delta V = negative \) 

Moving opposite to direction of current  \(\Delta V = positive\)

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For Battery

Moving in the direction of emf (+ve to -ve)

\(\Delta V = -\mathcal{E}\)

Moving opposite to direction of emf (-ve to +ve)

\(\Delta V = +\mathcal{E}\)

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For loop 1

\(\mathcal{E} _1- I_1R_2 - \mathcal{E}_2 - IR_1 = 0\)

(Traversing clockwise)

image image

For loop 2

\(-(I-I_1)R_3 + R_2I_1 = 0\)

(Traversing clockwise)

image image

Which option is correct by Kirchhoff's second law for the given circuit? \(\mathcal{E}_1>\mathcal{E}_2\)

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A

For loop \(I\)

\(\mathcal{E} _1- I_1R_2 + \mathcal{E}_2 + I_1R_1 = 0\)

For loop \(II\)

\(\{-(I - I_1) R_3 \} + I_1R_2 = 0\)

.

B

For loop \(I\)

\(\mathcal{E} _1+ IR_2 + \mathcal{E}_2 + I_1R_1 = 0\)

For loop \(II\)

\(\{-(I - I_1) R_3 \} + I_1R_2 = 0\)

C

For loop \(I\)

\(\mathcal{E} _1- I_1R_2 - \mathcal{E}_2 - IR_1 = 0\)

For loop \(II\)

\(\{-(I - I_1) R_3 \} + I_1R_2 = 0\)

D

For loop \(I\)

\(\mathcal{E} _1- I_1R_2 + \mathcal{E}_2 + I_1R_1 = 0\)

For loop \(II\)

\(\{-(I - I_1) R_3 \} - I_1R_2 = 0\)

Option C is Correct

Two loops having Common Connecting Wire

  • Consider two loops, i.e., loop \(I\) and loop \(II\) connected by a common connecting wire as shown in figure.
  • Let current \(I_1\) is flowing in loop \(I\)and current \(I_2\) in loop \(II\) .
  • Current \(I_1\) is flowing from battery  \(\mathcal{E}_1\) in loop \(I\) and again returning to the battery, i.e., the amount of charge flowing through battery \((\mathcal{E}_1)\) is same as amount of charge returning to battery \((\mathcal{E}_1)\).
  • Similarly, battery \((\mathcal{E}_2)\) is responsible for current  \(I_2\) in loop \(II\) .
  • Hence, no current will flow in connecting wire, i.e., \(I'=0\)

Illustration Questions

Which option is correct according to Kirchhoff's law ?

A For loop \(I\)  \(\mathcal{E}_1- I_1R_1=0\)  For loop \(II\)   \(\mathcal{E}_2- I_2R_2=0\)

B For loop \(I\)  \(\mathcal{E}_1+ I_1R_1=0\)  For loop \(II\)   \(-\mathcal{E}_2- I_2R_2=0\)

C For loop \(I\)  \(-\mathcal{E}_1- I_1R_1=0\)  For loop \(II\)  \(\mathcal{E}_2- I_2R_2=0\)

D For loop \(I\)  \(-\mathcal{E}_1+ I_1R_1=0\) For loop \(II\) \(-\mathcal{E}_2- I_2R_2=0\) 

×

For Resistor 

Moving in the direction of current \(\Delta V = negative\) 

Moving opposite to direction of current  \(\Delta V = positive\) 

image

For Battery 

Moving in the direction of emf (+ve to -ve)

\(\Delta V = -\mathcal{E}\)

Moving opposite to direction of emf (-ve to +ve)

\(\Delta V = +\mathcal{E}\)

image

For loop I   (Traversing clockwise)

 

image image

For loop II   (Traversing anticlockwise)

 

image image

Which option is correct according to Kirchhoff's law ?

image
A

For loop \(I\)

 \(\mathcal{E}_1- I_1R_1=0\)

 For loop \(II\)  

\(\mathcal{E}_2- I_2R_2=0\)

.

B

For loop \(I\)

 \(\mathcal{E}_1+ I_1R_1=0\)

 For loop \(II\)  

\(-\mathcal{E}_2- I_2R_2=0\)

C

For loop \(I\)

 \(-\mathcal{E}_1- I_1R_1=0\)

 For loop \(II\) 

\(\mathcal{E}_2- I_2R_2=0\)

D

For loop \(I\)

 \(-\mathcal{E}_1+ I_1R_1=0\)

For loop \(II\)

\(-\mathcal{E}_2- I_2R_2=0\) 

Option A is Correct

Illustration Questions

Calculate potential across point A and B.

A 10 V

B 2 V

C 5 V

D 4 V

×

Let current \(I_1\;,I_2 \) and \(I_3\) are flowing in the circuit , as shown in figure

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For loop \(I\)  (Traversing clockwise)

\(19V - 300 I_1 - 100 I_2- 12V = 0\)

or,   \(300I_1 + 100 I_2 =7V \;\;\;\;\;\)       ...(1)

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For loop \(II\)      (Traversing clockwise)

\(12 V + 100 I_2 - 200 I_3 = 0\)

or,   \( 200 I_3 - 100 I_2 = 12V\)            ...(2)

image

By Kirchhoff's law

\( I_1= I_2+ I_3 \)

or,  \( I_3 = I_1- I_2 \)               ..(iii)

 

From equation (ii) and (iii)

\( 200 (I_1- I_2) - 100 I_2 = 12\,V\)

or, \(200 I_1 - 200 I_2 - 100 I_2 = 12\,V\)

or, \(200 I_1 - 300 I_2 =12\,V\)          ...(iv)

Solving equation (i) and (iv)

\(I_1 = 0.030\,A\)

\(I_2 =- 0.020\,A\)

(negative sign indicate opposite direction)

Potential across \(100 \Omega \)  resistor

\(\Delta V_{100} = I_2 R \)

or, \(\Delta V_{100} =\) \( 0.020 × 100\)

or, \(\Delta V_{100} =\) \(2\,V \)

Potential across A and B

\(\Delta V_{AB} =\) \(12\,V - 2\,V\)

             = \(10 \,V\)

Calculate potential across point A and B.

image
A

10 V

.

B

2 V

C

5 V

D

4 V

Option A is Correct

Short Circuit 

  • Two points are said to be short circuited when they are connected together by a conducting wire.
  • The resistance of conducting wire is assumed to be negligible (zero).   
  • In this circuit , current across resistance is given as

            Current across resistance \(R_1 = I\) 

           Current across resistance \(R_2 = I_1\)

            Current across resistance \(R_3 = I- I_1\) 

 

  • When resistance \(R_3\) is short circuited, the potential of point Q and point S will be same.

            So, \(\Delta V_{R_3} = 0\)

  • The potential across parallel connection is same.

            So, \(\Delta V_{R_2} = 0\)  

  • Hence, no current will be flowing in \(R_2\) and \(R_3\) . The circuit reduces to

  • Applying Kirchhoff's second law

           \(\mathcal{E}_1 - IR_1=0\)

           \(\Rightarrow \mathcal{E}_1 = IR_1\)

             \(\Rightarrow I = \dfrac {\mathcal{E}_1}{R_1}\)

Illustration Questions

Calculate the value of \(I_1 , I_2\) and \(I_3\) in the following circuit, if \(2 \Omega\) resistance is short circuited. 

A \(2.5\,A\)

B \(2\, A\)

C \(4 \,A\)

D \(5 \,A\)

×

Since , current across short circuit is always zero.

So, \(I_3 =0\)     [ \(\therefore\) \(\Delta V_{2\Omega} = 0\)

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Since, in parallel connection voltage is same.

So, \(\Delta V_{5\Omega} = \Delta V_{2\Omega}\)

Also, \(I_2 = 0\)

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By Kirchhoff's second law for remaining circuit 

\(10 - 4 I_1 = 0\)

\(\Rightarrow I_1 = \dfrac {10}{4}\)

\(I_1 = 2.5\,A\)

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Calculate the value of \(I_1 , I_2\) and \(I_3\) in the following circuit, if \(2 \Omega\) resistance is short circuited. 

image
A

\(2.5\,A\)

.

B

\(2\, A\)

C

\(4 \,A\)

D

\(5 \,A\)

Option A is Correct

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