The current passing through a resistor r across which a potential difference V is applied is given by I = V/r.
When cells are connected in series, the combined voltage of the battery of cells is the sum of the individual voltage of each cell. If the four 1.5 V cells that are available are connected in series, the net voltage that is obtained is 1.5*4 = 6 V.
Each of the cells has an internal resistance of 0.2 ohm. When they are joined in series, the internal resistance of the cells can also considered to be connected in series. The equivalent resistance of a resistor formed by connecting several resistors in series is equal to the sum of the resistances of the resistors. This gives the total internal resistance as 0.2*4 = 0.8 ohm
If the 10 ohm resistor is connected across the 6 V battery, the total resistance connected across the battery is 10.8 V.
This gives the current flowing across the 10 ohm resistor equal to 6/10.8 = 5/9 amp.
The maximum current that can be passed through the 10 ohm resistor is 5/9 A.
Is the 0.2 ohm for all 4 of the cells? Or does each cell have 0.2 ohm each? Since it doesn’t put ‘total internal resistance’, I’m going to go with each cell having 0.2 ohm of resistance each.
Total resistance of the circuit= 10 + (4 x 0.2) = 10.8 ohm
Total voltage of the circuit = 4 x 1.5 = 6.00 V
And going by the formula, Voltage = Current x Resistance (V=IR),
Current = Voltage / Resistance (I=V/R)
Current= 6.00 / 10.8 = 0.556 A (rounding off to 3 significant figures)