Question

If n cells each of emp $r_{1}$ and internal resistance r are connected in parallel, then the total emf and internal resistance will be

• A. $r_{2}$,$r_{1}$
• B. $r_{2}$,nr
• C. n $\rho$,$\frac{r_{1}}{r_{2}}\frac{\rho}{2}$
• D. n$\frac{r_{2} - r_{1}}{r_{1}r_{2}}\frac{\rho}{4\pi}$, nr

Answer 1

Answer: (A)

$r_{2}$,$r_{1}$

Answer 2

: In the parallel combination.

$\frac{\varepsilon_{eq}}{r_{eq}}\frac{\varepsilon 1}{r_{1}} + \frac{\varepsilon_{2}}{r_{2}} + ..... + \frac{\varepsilon n}{r_{n}}$

$\frac{4}{r_{eq}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + ... + \frac{1}{r_{n}}$

($\because\varepsilon_{1} = \varepsilon_{2} = \varepsilon_{3} = .... = \varepsilon_{n} = \varepsilon$ and $r_{1} = r_{2} = r_{3} = ....r_{n} = r$)

$\therefore\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon}{r} + \frac{\varepsilon}{r} + .... + \frac{\varepsilon}{r} = n\frac{\varepsilon}{r}$ …(i)

$\frac{1}{r_{eq}} = \frac{1}{r} + \frac{1}{r} + .... + \frac{1}{r} = \frac{n}{r}$

$r_{eq} = r/n$ …(ii)

From (i) and (ii),

$\varepsilon_{eq} = n\frac{\varepsilon}{r} \times r_{eq} = n \times \frac{\varepsilon}{r} \times \frac{r}{n} = \varepsilon$