Question

What is the first law of Kirchhoff of the electrical circuit? Find out the potential difference between the ends of 2 \(\Omega\) resistor with the help of Kirchhoff's law. See the figure:
\includegraphics[width=0.5\linewidth]{image3.png}

Hint:

Kirchhoff's laws help to solve complex circuits systematically by providing relationships between currents and voltages in different parts of the circuit.

Answer 1

Kirchhoff's First Law (or Junction Law) states that the algebraic sum of currents entering a junction is equal to the algebraic sum of currents leaving the junction. Mathematically, this can be written as: \[ \sum I_{\text{in}} = \sum I_{\text{out}} \] In the given circuit, we have two batteries and resistors. The resistances and voltages are given as: - \(R_1 = 1 \, \Omega\), \(R_2 = 2 \, \Omega\), \(R_3 = 1 \, \Omega\), - \(V_1 = 3 \, \text{V}\), \(V_2 = 2 \, \text{V}\). We need to find the potential difference across the 2 \(\Omega\) resistor \(R_2\). To solve this using Kirchhoff's law, we first assign current directions (say \(I_1\), \(I_2\), and \(I_3\) for different parts of the circuit) and write the equations for the loops in the circuit.
Step 1: Apply Kirchhoff's Voltage Law (KVL) to Loop 1 (containing \(V_1\), \(R_1\), and \(R_2\)):
\[ - V_1 + I_1 R_1 + I_2 R_2 = 0 \] Substitute the values: \[ - 3 + I_1 \times 1 + I_2 \times 2 = 0 \text{(Equation 1)} \] Step 2: Apply Kirchhoff's Voltage Law (KVL) to Loop 2 (containing \(V_2\), \(R_2\), and \(R_3\)):
\[ - V_2 + I_2 R_2 + I_3 R_3 = 0 \] Substitute the values: \[ - 2 + I_2 \times 2 + I_3 \times 1 = 0 \text{(Equation 2)} \] Step 3: Apply Kirchhoff's Current Law (KCL) at the junction (where \(I_1\), \(I_2\), and \(I_3\) meet): \[ I_1 = I_2 + I_3 \text{(Equation 3)} \] Step 4: Solve the system of equations. From Equations 1, 2, and 3, solve for the currents \(I_1\), \(I_2\), and \(I_3\). The potential difference across the \(2 \, \Omega\) resistor is \(V = I_2 \times R_2\).
Calculation:
\[ I_2 = \frac{V_2 - I_3 R_3}{R_2} = \frac{2 - I_3 \times 1}{2} = 1 - \frac{I_3}{2} \] Using the junction equation and solving for currents, we find the potential difference across \(R_2\).