Two cells one of emf 1.4 V and internal resistance `0.6 Omega` the other of emf 2.5. V and internal resistance `0.3 Omega` are connected in parallel and the combination is connected in series with an external resistance of `4 Omega`. What is the current through this resistance?

Explanation

Suppse in fig.1.40 the potential difference between A and B , `V_A-V_B=V`.

So, for the first cell `V=E_1-I_1r_1or,I_1=(E_1-V)/r_1`
Similarly for the second cell `I_2=(E_2-V)/r_2`
For the resistance R of the external circuit `I=V/R`
Clearly, `I=I_1+I_2 or,V/R=(E_1-V)/r_1+(E_2-V)/r_2`
or,`V(1/R+1/r_1+1/r_2)=E_1/r_1+E_2/r_2`
or,`V(1/4+1/0.6+1/0.3)=1.4/0.6+2.5/0.3`
or,`V(0.25+1.66+3.33)=2.33+8.33`
or,`V=10.66/5.24V`
`therefore` Current through R,
`I=V/R=10.66/(5.24times4)=0.508A`