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Question
Two cells of emf E1, E2 and internal resistance r1 and r2 respectively are connected in parallel as shown in the figure.
Deduce the expressions for
(1) the equivalent e.m.f of the combination
(2) the equivalent resistance of the combination, and
(3) the potential difference between the point A and B.
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Solution
Let I1 and I2 be the current in two cells with emf E1 and E2 and internal resistance r1 and r2
So I = I1 + I2
Now let V be the potential difference between the points A and B. Since the first cell is connected between the points A and B.
V = potential difference across first cell
V = E1 − I1r1
`or I_1 =(E_1-V)/r_1`
Now, the second cell is also connected between the points A and B. So,
`or I_2 =(E_2-V)/r_2`
Thus, substituting for I1 and I2
`I =(E_1-V)/r_1 + (E_2-V)/r_2`
`or, I (E_1 /r_1 + E_2 /r_2)-V(1 /r_1 +1 /r_2)`
`V =((E_1r_2 + E_2r_1)/(r_1+r_2)) -I((r_1r_2)/(r_1 +r_2)) ..... (1)`
If E is effective e.m.f and r, the effective internal resistance of the parallel combination of the two cells then V = E − Ir …..(2)
Comparing (1) and (2)
`(1) E= (E_1r_2 + E_2r_1)/(r_1+r_2)`
This is equivalent e.m.f of the combination
`(2)r =(r_1r_2)/(r_1 +r_2)`
This is equivalent resistance of the combination.
(3) The potential difference between the point A and B is
V = E − Ir
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