A 4.5 V flash light battery gives a current of 0.5 A , if connected across a resistance of 7.5 ohm.

Question

A 4.5 V flash light battery gives a current of 0.5 A , if connected across a resistance of 7.5 ohm. Then the short circuit current is\

Professor PC Thomas Classes · Answer

$I=\frac{E}{R+r}$\ $0.5=\frac{4.5}{7.5+\mathrm{r}}$\ Solving $\mathrm{r}=1.5 \Omega$\ Short circuiting means putting $\mathrm{R}=0$, ie., connecting + ve and - ve by a wire of zero resistance.\ $	herefore \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}}=\frac{4.5}{1.5}=3 \mathrm{~A}$\