# NEET Physics Current Electricity Questions Solved

PMT - 2005

Two cells of equal e.m.f. and of internal resistances r1 and ${r}_{2}\left({r}_{1}>{r}_{2}\right)$ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be

(1) ${r}_{1}+{r}_{2}$

(2) ${r}_{1}-{r}_{2}$

(3) $\frac{{r}_{1}+{r}_{2}}{2}$

(4) $\frac{{r}_{1}-{r}_{2}}{2}$

(2) Let the voltage across any one cell is V, then

$V=E-ir=E-{r}_{1}\text{\hspace{0.17em}}\left(\frac{2E}{{r}_{1}+{r}_{2}+R}\right)$

But V = 0

$E-\frac{2E{r}_{1}}{{r}_{1}+{r}_{2}+R}=0$

${r}_{1}+{r}_{2}+R=2{r}_{1}$

$R={r}_{1}-{r}_{2}$

Difficulty Level:

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