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.

- 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.

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

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

- 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.

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

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

- 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.

\(\Delta V = IR\)

A Voltmeter reading across \( R_1\)

B \(R_4 I_6\)

C \(R_3 I_5\)

D None of these

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

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

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

- 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., \(I'=0\)

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

- 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

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

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

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

A \(2.5\,A\)

B \(2\, A\)

C \(4 \,A\)

D \(5 \,A\)