Physics - Current Electricity Question with Solution | TestHub

${\mathrm{R}}_{\mathrm{G}}=60.00\Omega ,$ shunt resistance, ${\mathrm{r}}_{\mathrm{s}}=0.02\Omega$ Total resistance in the circuit is ${\mathrm{R}}_{\mathrm{G}}+3=63\Omega$ Hence, $\mathrm{I}=3/63=0.048\mathrm{A}$ Resistance of the galvanometer converted to an ammeter is, $$\frac{{\mathrm{R}}_{\mathrm{G}}{\mathrm{r}}_{\mathrm{s}}}{{\mathrm{R}}_{\mathrm{G}}+{\mathrm{r}}_{\mathrm{s}}}=\frac{60\Omega \times 0.02\Omega }{(60+0.02)\Omega }=0.02\Omega$$ Total resistance in the circuit $=0.02+3$ $$=3.02\Omega$$ Hence, $\mathrm{I}=3/3.02=0.99\mathrm{A}$