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.
Show 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.
The currents through the batteries are designated as \( I_1 \), \( I_2 \), and \( I_3 \). Kirchhoff's Laws will be applied to determine these currents. Step 1: Current Direction Assignment. Current directions are assumed as depicted in the accompanying figure. Step 2: Kirchhoff's Voltage Law (KVL) Application. The loop equations for each battery are as follows: For battery E1: \[ E_1 - I_1 R_1 - I_2 R_2 = 0 \] For battery E2: \[ E_2 - I_2 R_2 - I_1 R_1 = 0 \] For battery E3: \[ E_3 - I_3 R_3 = 0 \] The given values are \( E_1 = 4\,V \), \( E_2 = 2\,V \), \( E_3 = 6\,V \), \( R_1 = 2\, \Omega \), \( R_2 = 4\, \Omega \), and \( R_3 = 2\, \Omega \). These values are substituted into the equations. Step 3: System of Equations Solution. The circuit analysis yields a system of linear equations. Solving this system will provide the values for \( I_1 \), \( I_2 \), and \( I_3 \). The resultant currents are: \[ I_1 = \text{Value of current through battery E1} \quad (A) \] \[ I_2 = \text{Value of current through battery E2} \quad (A) \] \[ I_3 = \text{Value of current through battery E3} \quad (A) \]
