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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

chand , 7 Years ago

Grade 12

2 Answers

Arun

Last Activity: 7 Years ago

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.

Kushagra Madhukar

Last Activity: 4 Years ago

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

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