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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
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.
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) + r1r2which is the required expression for the current through the load Thanks and regards,Kushagra
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