Question

State Kirchhoff’s laws. Apply these laws to find the values of current flowing in the three branches of the given circuit.

Hint:

When using Kirchhoff's Laws, make sure to carefully assign current directions and carefully apply the loop and junction rules. Solving the system of equations will give the correct values for currents in each branch of the circuit.

Answer 1

The circuit consists of: - A 5 V battery in series with a 2Ω resistor. - A 10 V battery in series with a 1Ω resistor and a 2Ω resistor. Step 1: Branch Current Assignment: Assign \( I_1 \) to the current in the branch with the 5V source and 2Ω resistor. Assign \( I_2 \) to the current in the branch with the 10V source and 1Ω resistor. Assign \( I_3 \) to the current in the branch with the 2Ω resistor. Step 2: Application of Kirchhoff's Laws: For loop 1 (5V source, 2Ω resistor): \[ 5 - 2I_1 = 0 \quad \Rightarrow \quad I_1 = \frac{5}{2} \, \text{A} \] For loop 2 (10V source, 1Ω resistor, 2Ω resistor): \[ 10 - 1I_2 - 2I_3 = 0 \] At the junction: \[ I_1 = I_2 + I_3 \] Substituting \( I_1 = \frac{5}{2} \): \[ \frac{5}{2} = I_2 + I_3 \] Step 3: System of Equations Solution: The system of equations is: 1. \( 10 - I_2 - 2I_3 = 0 \) 2. \( \frac{5}{2} = I_2 + I_3 \) From equation 1, solve for \( I_2 \): \[ I_2 = 10 - 2I_3 \] Substitute this into equation 2: \[ \frac{5}{2} = (10 - 2I_3) + I_3 \] \[ \frac{5}{2} = 10 - I_3 \] \[ I_3 = 10 - \frac{5}{2} = \frac{15}{2} \, \text{A} \] Substitute \( I_3 = \frac{15}{2} \) back into the expression for \( I_2 \): \[ I_2 = 10 - 2 \times \frac{15}{2} = 10 - 15 = -5 \, \text{A} \] The calculated current values are: - \( I_1 = \frac{5}{2} \, \text{A} \) - \( I_2 = -5 \, \text{A} \) - \( I_3 = \frac{15}{2} \, \text{A} \) The negative value for \( I_2 \) indicates that the actual current direction in the second branch is opposite to the assumed direction.