Physics - Current Electricity Question with Solution | TestHub

Here, emf of the battery$=8 \mathrm{V}$

The internal resistance of the battery,$r=0.5 \mathrm{\Omega }$

The voltage of D. C supply =$120 \mathrm{V}$

External resistance,$R=15.5 \mathrm{\Omega }$

When the battery of emf$8 \mathrm{V}$is charged from a$\mathrm{DC}$supply of$120 \mathrm{V}$, the effective emf in the circular is

$\epsilon =120-8=112 V$

The total resistance of the circuit$=R+r$

$=\left(15.5+0.5\right) \mathrm{\Omega }=16 \mathrm{\Omega }$

$\therefore$Current in the circuit during charging

$I=\frac{\epsilon }{R+r}=\frac{112}{16}=7\mathrm{ }\mathrm{A}$

$\therefore$The terminal voltage of the battery,

$V=\mathrm{e}\mathrm{m}\mathrm{f}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{b}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{y}+\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{b}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{y}$

$=8+Ir=8+7\left(0.5\right)=11.5\mathrm{ }\mathrm{V}$