Physics Chapter 1 Lesson 4 - Kirchhoff's Laws PDF

# Chapter 1: Lesson Four - Kirchhoff's Laws ## Kirchhoff's Laws Some electric circuits are too complicated to be analysed using Ohm's law. Therefore, the German scientist Kirchhoff set two laws that enable us to analyse such circuits. ### Kirchhoff's First Law - The electric current that passe...

# Chapter 1: Lesson Four - Kirchhoff's Laws ## Kirchhoff's Laws Some electric circuits are too complicated to be analysed using Ohm's law. Therefore, the German scientist Kirchhoff set two laws that enable us to analyse such circuits. ### Kirchhoff's First Law - The electric current that passes through metallic conductors is a stream of negative free electrons (electric charges) flowing from one point to another. - According to the law of conservation of electric charge, the quantity of electric charge flowing into a node within a certain interval of time is equal to the quantity of electric charge flowing out of that node in the same interval of time. - The electric current intensity (in amperes) is the quantity of electric charges (in coulombs) crossing a certain cross-section in one second. - Kirchhoff deduced his first law (Kirchhoff's current law) as follows: **Kirchhoff's First Law**: At any node (junction) in a closed electrical circuit, the sum of currents entering the node is equal to the sum of currents leaving that node. **Or** The algebraic sum of the electric currents at a point (a node) in a closed circuit equals zero. ### Applying Kirchhoff's First Law 1. The sum of currents entering a node = The sum of currents leaving the node 2. The algebraic sum of the currents at a node in a closed circuit = Zero **The Mathematical Formula**: ΣI_in = ΣI_out ΣI = 0 **At the node (junction)** - The current flowing into the point and that flowing out of it have a positive sign (+). - The current flowing into the point has a positive sign (+). - The current flowing out of the point has a negative sign (-). **Example 1** In the opposite figure, calculate the current intensity (I) and determine its direction. **Solution** - Assume that the direction of I is into node C. - Based on Kirchhoff's first law: ΣI_in = ΣI_out 4 + 5 + 2 + I = 8 I = -3 A - The current intensity I equals 3 A and its direction is out of point C (opposite to the assumed direction). ## Kirchhoff's Second Law Previously, you have studied: **Potential Difference (V)** - It expresses the work done or the energy required to transfer a unit electric charge across a component (part of the circuit) in the circuit. - **Calculated from the relation**: V = IR Where: (R) is the resistance of the component in the circuit across which the potential difference is required to be calculated. **Electromotive Force (V_B)** - It expresses the work done or the energy required to transfer a unit electric charge once throughout the whole closed circuit. - **Calculated From the relation**: V_B = I (R + r) Where: (R) is the total external resistance, (r) is the internal resistance of the source. Accordingly, Kirchhoff formulated his second law to find a relation between the potential difference (V) and the electromotive force (VB) (Kirchhoff's voltage law) as follows: **Kirchhoff's Second Law**: The algebraic sum of the electromotive forces through any closed loop equals the algebraic sum of the potential differences across the components (resistors) of that loop. **Or** The algebraic sum of the potential differences through any closed loop equals zero. **Notes:** - When solving problems using Kirchhoff's second law, a direction has to be assumed for every closed loop either clockwise or counterclockwise. - Kirchhoff's second law is applied to any closed loop. - Kirchhoff's second law is considered as an application of the law of conservation of energy, which states that in any closed loop the electric energy that is produced from the source equals the consumed electric energy through the components of the loop. ## Applying Kirchhoff's Second Law 1. In a closed loop, the algebraic sum of the electromotive forces = The algebraic sum of the potential differences. 2. The algebraic sum of the potential differences in a closed loop = Zero **The Mathematical Formula**: ΣV_B = ΣIR ΣV = 0 **Rules to determine the signs of potential differences across resistors and batteries**: **(1) If the used mathematical formula is ΣV_B = ΣIR** - If the assumed direction is from the negative pole to the positive pole inside the source (battery), then the emf for this source takes a positive sign. - If the assumed direction is from the positive pole to the negative pole inside the source (battery), then the emf for this source takes a negative sign. **(2) If the used mathematical formula is ΣV = 0** - If the assumed direction is from the negative pole to the positive pole inside the source (battery), then the emf for this source takes a positive sign. - If the assumed direction is from the positive pole to the negative pole inside the source (battery), then the emf for this source takes a negative sign. **Example** In the circuit shown in the opposite figure, calculate each of V_z, V_y. **Solution** V_z = 4 V, V_y = 2 V, V_x =?, V_w = ? Assume the directions of the loops as shown in the circuit: - Applying Kirchhoff's second law on the loop (adcba): ΣV = 0 -V_z - V_w + V_x = 0 -V_z - 2 + 4 = 0 V_x = 2 V - Applying Kirchhoff's second law on the loop (bcdb): V_y - V_z = 0 V_y - 4 = 0 V_y = 4 V ## How to solve problems using Kirchhoff's Laws: For an electric circuit as that shown in the figure, to calculate the electric current intensity passing through each resistance, we go through the following steps: 1. **If some resistors are connected together either in parallel or series, it's preferred to find their equivalent resistance before applying Kirchhoff's laws.** 2. **Assume a certain direction for each unknown current (These directions are not all necessarily to be correct).** **Determine the number of the unknown quantities required to be calculated.** **Assume a direction for each closed loop randomly (clockwise or anti-clockwise).** 3. **Apply Kirchhoff's first law at a node for the current such that: ΣI_in = ΣI_out Thus, you obtain the first equation.** 4. **Choose a closed loop and apply Kirchhoff's second law (the sign rule should be considered) such that: ΣV_B = ΣIR. Thus, you obtain the second equation.** 5. **Repeat the previous step for multiple loops until the number of equations equals the number of the unknown values.** 6. **Solve equations ① using the calculator to obtain the unknown values which are: I₁ = 1.5A, I₂ = -0.5A, I₃ = 2 A.** 7. **If the calculated value of the current is: ** - **Positive:** then the correct direction of the current is the same as the direction you assumed at the beginning. - **Negative:** then the correct direction of the current is opposite to the direction you assumed at the beginning.