Current Electricity - Result Question 54
57. Two cells, having the same e.m.f., are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2(r_1>r_2)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is
[2006]
(a) $\frac{r_1+r_2}{2}$
(b) $\frac{r_1-r_2}{2}$
(c) $r_1+r_2$
(d) $r_1-r_2$
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Answer:
Correct Answer: 57. (d)
Solution:
- (d) Current in the circuit
$=\frac{E+E}{r_1+r_2+R}=\frac{2 E}{r_1+r_2+R}$
P.D. across first cell $=E-i r_1$
$=E-\frac{2 E \times r_1}{(r_1+r_2)+R}$
Now, $E=\frac{2 E r_1}{(r_1+r_2)+R}=0$
$\Rightarrow E=\frac{2 E r_1}{r_1+r_2+R} \Rightarrow 2 r_1=r_1+r_2+R$
$R=r_1-r_2$
In series grouping of cells their emf’s are additive or subtractive while their internal resistances are always additive. If dissimilar plates of cells are connected together their emf’s are added to each other while if their similar plates are connected together their emf’s are subtractive.
$ \begin{aligned} & E_1 \quad E_2 \\ & -\vec{\longmapsto} \quad-\longmapsto \longmapsto \\ & E _{e q}=E_1+E_2 \quad E_1 \quad E_2 \\ & r _{e q}=r_1+r_2 \quad E _{e q}=E_1-E_2(E_1>E_2) \\ & r _{e q}=r_1+r_2 \end{aligned} $
