Two cells of voltage 10 V and 2 V and internal resistances 10 Ω and 5 Ω respectively, are connected in parallel with the positive end of the 10 V battery connected to the negative pole of the 2V battery (figure). Find the effective voltage and effective resistance of the combination.

                                

Hint: Apply Kirchoff's laws.
Step 1: Applying Kirchhoff's junction rule, I1=I+I2
Applying Kirchhoff's IInd law/loop rule applied in outer loop containing 10 V cell and resistance R, we have;
10=IR+10I1                   ...i
Step 2: Applying Kirchhoff IInd law/loop rule applied in upper loop containing 2 V cell and resistance R, we have;
    2=5I2RI=5(I1I)RI 
or 4=10I110I2RI                                    ...ii
Solving Eqs. (i) and (ii),
 6=3RI+10I
      2=I(R+103)
Step 3: Also, the external resistance is R. Ohm's law states that;
V=I(R+R eff )
On comparing, we have V= 2 V and effective internal resistance;
Relf = 103 Ω
Since the effective internal resistance (Reff) of two cells is 103 Ω being the parallel combination of 5Ω and 10Ω. The equivalent circuit is given below;