# two cells of EMF E1 and E2 and internal resistance R1 and R2 are connected in parallel between the points A and B deduce the expression for the equivalent emf of the combination

Arun
25750 Points
6 years ago

Given,
Emf of first cell = E1
Emf of second cell = E
Internal resistances of two cell are r1 and r2 respectively.

Effective emf is given by, Eeff = E1r1 + E2 r2r1 + r2

Effective internal resistance, reff = r1 r2r1 + r2

Now, the total resistance of the circuit would be the sum of load resistance and effective internal resistance.

Rtotal = R + reff = R + r1r2r1+r2
Current through the load will be the same as the current through the circuit.
Thus,
I = EeffRtotal  = E1r1 + E2r2r1 +r2R+ r1r2r1 + r2 = E1r1 + E2 r2R(r1+r2) + r1r2

which is the required expression for the current through the load.
3 years ago
Dear student,

Given, Emf of first cell = E1
Emf of second cell = E2
Internal resistances of two cell are r1 and r2 respectively.
Effective emf is given by, Eeff = E1r1 + E2 r2r1 + r2
Effective internal resistance, reff = r1 r2r1 + r2
Now, the total resistance of the circuit would be the sum of load resistance and effective internal resistance.

Rtotal = R + reff = R + r1r2r1+r2
Current through the load will be the same as the current through the circuit.
Thus,
I = EeffRtotal  = E1r1 + E2r2r1 +r2R+ r1r2r1 + r2 = E1r1 + E2 r2R(r1+r2) + r1r2
which is the required expression for the current through the load

Thanks and regards,
Kushagra