Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 Ω . The internal resistance of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the loa

Question

Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 Ω . The internal resistance of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between,

TestHub · Accepted Answer

As shown in figure two cells are connected in parallel with the load resistance of $R = 10 Ω$ . Let current $i_{1}, i_{2}$ drawn from both cells as shown, from $K . C . L .$ $i = i_{1} + i_{2} . . . (1)$

Apply $K . V . L .$ for loop $A C D F$ ,

$E_{1} - i R - i_{1} r_{1} = 0 \Rightarrow 12 - 10 i - i_{1} = 0 . . . (2)$

From equations $(1)$ and $(2)$

$12 - 11 i_{1} - 10 i_{2} = 0 . . . (3)$

Now, for loop $B C D E$ ,

Apply $K . V . L .$

$E_{2} - 10 i - i_{2} r_{2} = 0 \Rightarrow 13 - 10 i - 2 i_{2} = 0 \Rightarrow (∵ i = i_{1} + i_{2}) ∴ 13 - 10 i_{1} - 12 i_{2} = 0 . . . (4)$

From equation $(3)$ and $(4)$ ,

$i_{1} = \frac{7}{16} A, i_{2} = \frac{23}{32} A$

So, net current in external load $R = 10 Ω$ is

$I = i_{1} + i_{2} = \frac{7}{16} + \frac{23}{32} = \frac{37}{32} A$

Then potential difference for a given load,

$V = I R = \frac{37}{32} \times 32 = 11 \cdot 56 V$

So, the potential difference lies between, $11 \cdot 5 V$ and $11 \cdot 6 V$ .

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