Kirchhoff's First Law

• Kirchhoff’s first law states that:

The sum of the?currents?entering a junction always equals the sum of the currents out of the junction

• A?junction is a point where at least three circuit paths meet

The current I into the junction is equal to the sum of the currents I₁, I₂?and I₃?out of the junction

• Kirchhoff’s first law is a consequence of?conservation of charge?
  • The charge is the same on both sides of the junction

• Kirchhoff’s first law is often written as follows:

∑I =?0

• Where:
  • ∑I?= net current entering and leaving the junction
  • The current entering a junction is defined as positive
  • The current leaving a junction is defined as negative

Step 1: Ammeter A₁?is in series with ammeter A

• • The current reading of ammeter A₁?is the same as that of ammeter A

A₁?= 0.270 A

Step 2: From Kirchhoff's first law, the total current entering the junction must be equal to the total current leaving the junction?

• • The current from ammeter A₁?enters a junction and splits into two branches
  • One branch has a current of 80 mA = 0.080 A
  • The current in the other branch is the reading of ammeter A₃

0.270 A = 0.080 A + A₃

A₃?= (0.270 – 0.080) A = 0.19 A

Step 3: Apply Kirchhoff's law to the other junction of the circuit?

• • The 0.080 A current enters a junction and splits into two branches
  • One branch has a current of 27 mA = 0.027 A
  • The current in the other branch is the reading of ammeter A₂

0.080 A = 0.027 A + A₂

A₂?= (0.080 – 0.027) A = 0.053 A

The current readings of the three ammeters are:

• • A₁?= 0.270 A = 270 mA
  • A₂?= 0.053 A = 53 mA
  • A₃?= 0.19 A = 190 mA

Kirchhoff's Second Law

• Kirchhoff’s second law states that:

The net?potential difference?in a closed loop is equal to zero?

• This is a consequence of?conservation of energy
  • The electric potential energy carried by the electrons as they flow through a circuit must be equal to the overall thermal energy transferred to the different components

The sum of the electric potential energies E₁?and E₂?provided by the two cells must be equal to the sum of the potential differences across the two fixed resistors

• Kirchhoff’s second law is often written as follows:

∑V =?0

• Where:
  • ∑V?= net potential difference in a closed loop
  • The potential difference entering a cell is defined as positive
  • The potential difference entering a resistor (or another circuit component offering resistance) is defined as negative

Step 1: Kirchhoff's second law states that the sum of all potential differences in the first loop is equal to zero?

• • The potential difference provided by the cell is positive and equal to 24 V
  • The potential difference across the first light bulb is negative and equal to 2V
  • The potential difference across the second light bulb is negative and equal to 12 V

V₁?= (24 – 2 – 12) V = 10 V

Step 2: Apply Kirchhoff's second law to the second loop of the circuit?

• • The potential difference across the fixed resistor is negative and equal to 8 V
  • The potential difference across each light bulb is negative and equal in value to the reading of V₂?(since they have the same resistance)
  • All these negative potential differences must be summed to:
    • The positive potential difference provided by the cell (24 V)
    • The negative potential difference across the light bulb connected in series with the cell (2 V)

2V₂?= (24 – 2 – 8) V = 14 V

V₂?= 7 V

Step 3: Apply Kirchhoff's second law to the third loop of the circuit

• • The reading of V₃?is equal to the sum of:
    • The positive potential difference provided by the cell (24 V)
    • The negative potential difference across the light bulb connected in series with the cell (2 V)

V₃?= (24 – 2) V = 22 V

The potential difference readings of the three voltmeters are:

• • V₁?= 10 V
  • V₂?= 7 V
  • V₃?= 22 V