Question

Six identical cells each of $e.m.f.2V$ and internal resistance $0.2\mathrm{\Omega }$ are connected in parallel to form a battery. An external resistance of $5\mathrm{\Omega }$ is connected to this battery. Draw the circuit diagram of the above arrangement. What is the current in $5\mathrm{\Omega }$ resistor.

Answer 1

Step-1 Given data & drawing a circuit diagram

$E.m.f.$ of each cell is $E=2V$

The number of cells is $n=6$

The internal resistance of each cell is $r=0.2\Omega$

The external resistance is $R=5\Omega$

Step-2 Finding equivalent resistance

Six identical cells are connected in parallel.

So, equivalent internal resistance

$$\begin{gathered} \frac{1}{{\mathrm{r}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{r}}_1}+\frac{1}{{\mathrm{r}}_2}+\frac{1}{{\mathrm{r}}_3}+........+\frac{1}{{\mathrm{r}}_6} \\ {\mathrm{r}}_{\mathrm{eq}}=\frac{\mathrm{r}}{\mathrm{n}}=\frac{0.2}{6}\mathrm{\Omega } \end{gathered}$$

Total resistance, ${\mathrm{R}}_{\mathrm{eq}}=\mathrm{external}\mathrm{resistance}+\mathrm{internal}\mathrm{resistance}$

$=(5+\frac{0.2}{6})\mathrm{\Omega }$

Step-3 Finding net $e.m.f.$

All cells are connected in parallel combination

So, net $e.m.f.$of cell is

$$\begin{gathered} E_{net}(\frac{1}{{\mathrm{r}}_1}+\frac{1}{{\mathrm{r}}_2}+\frac{1}{{\mathrm{r}}_3}+........+\frac{1}{{\mathrm{r}}_6})=\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}+......+\frac{E_6}{r_6} \\ {E}_{net}\left(\frac{1}{0.2}+\frac{1}{0.2}+\frac{1}{0.2}+........+\frac{1}{0.2}\right)=\frac{2}{0.2}+\frac{2}{0.2}+\frac{2}{0.2}+......+\frac{2}{0.2} \\ {E}_{net}\left(\frac{6}{0.2}\right)=\frac{6\times 2}{0.2} \\ {E}_{net}=2V \end{gathered}$$

Step-4 Finding current

Current through $5\mathrm{\Omega }$ resistance is

$$\begin{gathered} \mathrm{I}=\frac{Nete.m.f.ofcell}{{\mathrm{R}}_{\mathrm{eq}}} \\ I=\frac{E_{net}}{{\mathrm{R}}_{\mathrm{eq}}} \end{gathered}$$

$=\frac{2}{5+{\displaystyle \frac{0.2}{6}}}$

$=\frac{12}{30.2}$

$\mathrm{I}=0.39\mathrm{Ampere}.$

Hence, the current through $5\mathrm{\Omega }$ resistance is $0.39\mathrm{A}.$