Question 554: Current Electricity

Question

Two cells $$E_1$$ and $$E_2$$ having equal EMF '$$E$$' 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 $$\mathrm{E}_1$$ becomes zero. The value of '$$R$$' will be

Accepted Answer

The total current in the circuit is
$$\mathrm{I}=\frac{2 \mathrm{E}}{\mathrm{r}_1+\mathrm{r}_2+\mathrm{R}}\quad$$ ... (1) .... (Given cells are in series, $$\mathrm{E}+\mathrm{E}=2 \mathrm{E})$$
Now the potential drop across the first cell is $$V_1=E-r_1=0$$
$$\begin{aligned}
	herefore \quad & E-\left(\frac{2 E}{r_1+r_2+R}ight) 	imes r_1=0 \
& \frac{2 E}{r_1+r_2+R}=\frac{E}{r_1} \Rightarrow 2 r_1=r_1+r_2+R \
	herefore \quad & R=r_1-r_2
\end{aligned}$$

Answer

$$r_1-r_2$$

Answer

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

Answer

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

Answer

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

Current Electricity

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