Informative line

### 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.

$$\sum I _{Enter} = \sum I _{Leave}$$

$$I_{1} = I_{2}+I_{3}$$

#### 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.

junction at point P

$$\sum I _{Enter} = \sum I _{Leave}$$

$$\sum I _{Enter} = I = 5\,A$$

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

$$I_{2} +3 = 5$$

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

$$\Rightarrow I_{2} = 2\,A$$

### 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

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.

$$\sum I _{Enter} = \sum I _{Leave}$$

$$I_{1} = I_{2}+I_{3}$$

#### 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

$$\sum I _{Enter} = \sum I _{Leave}$$

$$I_{t} = I_{1} + I_ {X}$$

$$2\,A = 0.75\,A+ I_{X}$$

$$I_{X} = 2\,A - 0.75\,A$$

$$I_{X} = 1.25\,A$$

At Junction Y

$$\sum I_{ENTER} =\sum I_{LEAVE}$$

$$I_{X} = I_{2}+I_{3}$$

$$1.25 = 0.25 + I_{3}$$

$$I_{3} = 1.25 -0.25$$

$$I_{3} = 1\,A$$

### 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

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

$$\Delta V = -\mathcal{E}$$

#### 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  $$\Delta V = positive$$

For Battery

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

$$\Delta V = -\mathcal{E}$$

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

$$\Delta V = +\mathcal{E}$$

For loop 1

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

(Traversing clockwise)

For loop 2

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

(Traversing clockwise)

### 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 $$II$$

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

Option C is Correct

#### 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

For loop $$I$$  (Traversing clockwise)

$$19V - 300 I_1 - 100 I_2- 12V = 0$$

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

For loop $$II$$      (Traversing clockwise)

$$12 V + 100 I_2 - 200 I_3 = 0$$

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

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.

A

10 V

.

B

2 V

C

5 V

D

4 V

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.

$$\Delta V = IR$$

#### 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

$$\Delta V_1 = I_1\;R_1$$

Voltmeter reading across R2

$$\Delta V_2 = I_2\;R_2$$

Voltmeter reading across R3

$$\Delta V_3 = I_5\;R_3$$

Voltmeter reading across R4

$$\Delta V_4 = I_6\;R_4$$

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

$$\Delta V_3 = \Delta V_4$$

$$or,\,\, I_5\;R_3 = I_6\;R_4$$

### 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

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

$$\Delta V = -\,\mathcal{E}$$

#### 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  $$\Delta V = positive$$

For Battery

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

$$\Delta V = -\mathcal{E}$$

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

$$\Delta V = +\mathcal{E}$$

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

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

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

### 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

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

Option A 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., $$I'=0$$

#### 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  $$\Delta V = positive$$

For Battery

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

$$\Delta V = -\mathcal{E}$$

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

$$\Delta V = +\mathcal{E}$$

For loop I   (Traversing clockwise)

For loop II   (Traversing anticlockwise)

### 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 $$II$$

$$-\mathcal{E}_2- I_2R_2=0$$

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

$$\mathcal{E}_1 - IR_1=0$$

$$\Rightarrow \mathcal{E}_1 = IR_1$$

$$\Rightarrow I = \dfrac {\mathcal{E}_1}{R_1}$$

#### 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, $$I_3 =0$$     [ $$\therefore$$ $$\Delta V_{2\Omega} = 0$$

Since, in parallel connection voltage is same.

So, $$\Delta V_{5\Omega} = \Delta V_{2\Omega}$$

Also, $$I_2 = 0$$

By Kirchhoff's second law for remaining circuit

$$10 - 4 I_1 = 0$$

$$\Rightarrow I_1 = \dfrac {10}{4}$$

$$I_1 = 2.5\,A$$

### 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

$$5 \,A$$

Option A is Correct