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+2 votes
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in Physics by (150k points)
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Two cells of emfs ε1,ε2 and internal resistance r1 and r2 respectively are connected in parallel as shown in the figure. Deduce the expressions for 

(i) the equivalent e.m.f. of the combination, 

(ii) the equivalent resistance of the combination, and 

(iii) the potential difference between the points A and B. 

2 Answers

+1 vote
by (17.0k points)
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Best answer

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 \(I_1 = \frac{E_1 - V}{r_1}\)

Now, the second cell is also connected between the points A and B. So,

or \(I_2 = \frac{E_2 - V}{r_2}\)

Thus, substituting for I1 and I2

\(I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2}\)

or \(I\left(\frac{E_1}{r_1} + \frac{E_2}{r_2}\right) - V\left(\frac 1{r_1} + \frac 1{r_2}\right)\)

\(V = \left(\frac{E_1r_2 + E_2 r_1}{r_1 + r_2}\right) - I\left(\frac{r_1r_2}{r_1 + r_2}\right)\)    ......(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),

(i) \(E= \frac{E_1r_2 + E_2 r_1}{r_1 + r_2}\)

This is equivalent e.m.f of the combination

(ii) \(r = \frac{r_1r_2}{r_1 + r_2}\)

This is equivalent resistance of the combination.

(iii) The potential difference between the point A and B is

V = E − Ir

+2 votes
by (87.2k points)

Here, I = I1 + I2 ...(i)
Let V = Potential difference between A and B.
For cell ε1

Comparing the above equation with the equivalent circuit of emf ‘εeq’ and internal resistance ‘req’ then,

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