Physics - Current Electricity Question with Solution | TestHub

As shown in figure two cells are connected in parallel with the load resistance of $R=10\Omega$. Let current $i_1,i_2$ drawn from both cells as shown, from $K.\mathrm{C}.L.$ $i=i_1+i_2...\left(1\right)$

Apply $K.\mathrm{V}.L.$ for loop $A\mathrm{C}D\text{F}$,

$$\begin{gathered} E_1-iR-i_1{r}_1=0 \\ \Rightarrow 12-10i-i_1=\mathrm{0...}\left(2\right) \end{gathered}$$

From equations $\left(1\right)$ and $\left(2\right)$

$12-11i_1-10i_2=\mathrm{0...}\left(3\right)$

Now, for loop $B\mathrm{C}DE$,

Apply $K.\mathrm{V}.L.$

$$\begin{gathered} E_2-10i-i_2{r}_2=0 \\ \Rightarrow 13-10i-2i_2=0 \\ \Rightarrow \left(\because i=i_1+i_2\right) \\ \therefore 13-10i_1-12i_2=\mathrm{0...}\left(4\right) \end{gathered}$$

From equation $\left(3\right)$ and $\left(4\right)$,

$i_1=\frac{7}{16}\text{A},i_2=\frac{23}{32}\text{A}$

So, net current in external load $R=10\Omega$ is

$I=i_1+i_2=\frac{7}{16}+\frac{23}{32}=\frac{37}{32}\text{A}$Then potential difference for a given load,

$V=IR=\frac{37}{32}\times 32=11\cdot 56\mathrm{V}$

So, the potential difference lies between, $11\cdot 5\mathrm{V}$ and $11\cdot 6\mathrm{V}$.