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Grade: 12

                        

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

3 years ago

Answers : (2)

Arun
24742 Points
							
 
Given,
Emf of first cell = E1 
Emf of second cell = E
Internal resistances of two cell are r1 and r2 respectively. 
Load resistance = R 

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
Kushagra Madhukar
askIITians Faculty
605 Points
							
Dear student,
Please find the answer to your question.
 
Given, Emf of first cell = E1 
Emf of second cell = E2 
Internal resistances of two cell are r1 and r2 respectively. 
Load resistance = R 
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
2 months ago
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