Numerical Problems Based on Kirchhoff’s Laws for Class 12 Physics

1. Junction Rule (KCL):
Identify a principal junction (e.g., node P or R) and apply Kirchhoff’s Current Law. The algebraic sum of currents meeting at the junction is zero, providing an equation that relates $I_1, I_2$, and $I_3$.

2. Loop Rule (KVL):
Apply Kirchhoff’s Voltage Law to the specified closed loops:

• Loop PRSP: Traverse the loop, summing the potential changes across resistors ($-IR$) and batteries (using the sign of the exiting terminal). Equate the sum to zero.
• Loop PRQP: Traverse this second loop in a chosen direction, similarly summing the potential drops and gains, and equate to zero.

3. Solving the System:
Substitute the KCL equation into the two KVL equations to reduce the system to two variables. Solve the simultaneous linear equations to find the precise values of the currents.

Establish the current distribution across the network using Kirchhoff’s Current Law (KCL) at the nodes, expressing branch currents in terms of $I_1$ and another variable.

Select a closed loop containing the branch with current $I_1$. Apply Kirchhoff’s Voltage Law (KVL), maintaining consistent sign conventions for EMFs and resistive voltage drops.

Solve the resulting algebraic equations. The solution yields:

Note: The negative sign indicates that the actual direction of the current $I_1$ is opposite to the direction initially assumed in the circuit diagram.

1. Formulate Junction Equation:
Apply KCL at the main junction to relate the three currents. Depending on the labeled directions, this typically takes the form of $I_3 = I_1 + I_2$ or similar.

2. Formulate Loop Equations:
Apply KVL to two independent loops in the network to generate two linear equations involving $I_1, I_2$, and $I_3$.

3. Calculate Currents:
Solve the system of three equations. The evaluation resolves to:

The negative sign for $I_2$ indicates it flows opposite to its designated direction in the diagram.

Let $I_1$ be the current leaving the 4 V cell and $I_2$ be the current leaving the 2 V cell. By KCL, the current flowing through the external $8 \Omega$ resistor is $(I_1 + I_2)$.

Apply KVL to the loop containing the 4 V cell and the $8 \Omega$ resistor:

Apply KVL to the loop containing the 2 V cell and the $8 \Omega$ resistor:

Solve the simultaneous equations:
Multiply (1) by 5 and (2) by 4:

Subtracting the second from the first:

Substitute $I_1$ back into equation (1):

The current through the 4 V cell is $\frac{2}{3}\text{ A}$ (discharging), and the current through the 2 V cell is $\frac{1}{3}\text{ A}$ in the reverse direction (charging).