Step 1:

Two cells are connected in the series.

The emf of the first cell is 1.5 V.

The internal resistance of the first cell is 1.2 .

The emf of the second cell is 1.1 V.

The internal resistance of the second cell is 0.7 .

Draw a circuit with above specifications :

 Step 2:

Find the equivalent resistance :

Resistor parallel.

Let is equivalent parallel resistance.

Step 3:

Find the current in the circuit :

From ohms law : .

The current in the circuit : .

Where is the emf of the first cell,

         is the emf of the first cell,

         is the internal resistance of the first cell,

         is the internal resistance of the second cell,

            is equivalent parallel resistance.

So the current in the circuit :

Current in the series loop is same.

Therefore, the current in each cell is .

 Step 4:

(3)

The terminal voltage of first cell :

Suppose that the battery with emf  and internal resistance supplies a current  through an external load resistor , then the potential difference across the battery is .

Substitute , and in .

\ \

The terminal voltage of second cell :

Suppose that the battery with emf  and internal resistance supplies a current  through an external load resistor , then the potential difference across the battery is .

Substitute , and in .

 Step 5:

The Terminal voltage of first cell is .

The Terminal voltage of second cell is .

The cells are connected in series .

The net voltage is .

The voltage in the parallel loop is .

The voltage in the parallel loop same.

So, the voltage across each coil is same.

Current through the coil which has resistance  :

Use ohms law : .

Current through the coil which has resistance  :

Use ohms law : .

Solution:

The current in each cell is .

The Terminal voltage of first cell is .

The Terminal voltage of second cell is .

Current through the first coil is .

Current through the second coil is .