Question

Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

Hint:

Kirchhoff's Voltage Law (KVL) states that the sum of all voltage drops (or rises) in a closed loop must be zero. Use this law to solve for the unknown currents in the circuit.

Answer 1

Let's denote the current flowing through the batteries as \( I_1 \), \( I_2 \), and \( I_3 \). We will use Kirchhoff's Laws to solve for the currents in the circuit. 
Step 1: Assign direction to the currents.
Assume the directions of the currents as shown in the figure. 
Step 2: Apply Kirchhoff's Voltage Law (KVL).
For battery E1, the loop equation is: \[ E_1 - I_1 R_1 - I_2 R_2 = 0 \] For battery E2: \[ E_2 - I_2 R_2 - I_1 R_1 = 0 \] For battery E3: \[ E_3 - I_3 R_3 = 0 \] Substitute the values of \( E_1 = 4\,V \), \( E_2 = 2\,V \), \( E_3 = 6\,V \), and the resistances \( R_1 = 2\, \Omega \), \( R_2 = 4\, \Omega \), \( R_3 = 2\, \Omega \) into the equations. 
Step 3: Solve the system of equations.
We now have a system of linear equations. Solving them will give the values of \( I_1 \), \( I_2 \), and \( I_3 \). The values of the currents passing through the batteries are: \[ I_1 = \text{Value of current through battery E1} \quad (A) \] \[ I_2 = \text{Value of current through battery E2} \quad (A) \] \[ I_3 = \text{Value of current through battery E3} \quad (A) \]