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Question
Two cells of emfs E1 and E2 and internal resistances r1 and r2 respectively are connected in parallel as shown in the figure. Deduce the expression for the
- equivalent emf of the combination
- equivalent internal resistance of the combination
- potential difference between the points A and B.
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Solution
Here, I = I1 + I2 ...(i)
Let V = Potential difference between A and B.
For cell ε1
Then, V = ε1 – I1r1
⇒ I1 = `(ε_1 - V)/r_1`
Similarly, for cell ε2 I2 = `(ε_2 - V)/r_2`
Putting these values in equation (i)
I = `(ε_1 - V)/r_1 + (ε_2 - V)/r_2`
or I = `(ε_1/r_1 + ε_2/r_2) -V (1/r_1 + 1/r_2)`
or V = `((ε_1r_2 + ε_2r_1)/(r_1 + r_2)) -I((r_1r_2)/(r_1 + r_2))` ...(ii)
Comparing the above equation with the equivalent circuit of emf ‘εeq’ and internal resistance ‘req’ then,
V = εeq – Ireq ...(iii)
Then
- εeq = `(ε_1r_2 + ε_2r_1)/(r_1 + r_2)`
- req = `(r_1r_2)/(r_1 + r_2)`
- The potential difference between A and B
V = εeq – Ireq
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