Question 510: Current Electricity

Question

Two cells $E_1$ and $E_2$  having equal e.m.f ' $E$ ' and internal resistances $r_1$ and $r_2\left(r_1>r_2\right)$ respectively are connected in series. This combination is connected to an external resistance ' R '. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

Accepted Answer

Current in the circuit: $I=\frac{2 E}{R+r_1+r_2}$
Terminal p.d across $1^{	ext {st }}$ cell is $\mathrm{V}_1=\mathrm{E}-\mathrm{Ir}_1$
Given: $\mathrm{V}_1=0$
$$ \begin{aligned} \Rightarrow \quad & E-I r_1=0 \ & E-\left(\frac{2 E}{R+r_1+r_2}ight) r_1=0 \ & E=\frac{2 E r_1}{R+r_1+r_2} \ & R+r_1+r_2=2 r_1 \ \Rightarrow \quad & R=r_1-r_2 \end{aligned} $$

Answer

$\mathrm{r}_1-\mathrm{r}_2$

Answer

$r_1+r_2$

Answer

$\frac{r_1-r_2}{2}$

Answer

$\frac{r_1+r_2}{2}$

Current Electricity

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