Physics - Current Electricity Question with Solution | TestHub

$\frac{R}{\mathcal{l}}=0.5 \mathrm{\Omega } {\mathrm{m}}^{-1}$,$\mathrm{r}=1\mathrm{ }\mathrm{m}$

Length of semicircle$=\mathrm{\pi }r$

Hence the resistance of semicircle

$R=0.5\times \mathrm{\pi }r$

$R=\frac{1}{2}\times \mathrm{\pi }\times 1$

$R=\frac{\mathrm{\pi }}{2} \mathrm{\Omega }$

Resistance of diameter

$R=0.5\times 2r$

$R=0.5\times 2\times 1$

$R=1$



$\frac{1}{R_{\mathrm{e}\mathrm{q}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R_{\mathrm{e}\mathrm{q}}}=\frac{2}{\mathrm{\pi }}+\frac{1}{1}+\frac{2}{\mathrm{\pi }}$

$\frac{1}{R_{\mathrm{e}\mathrm{q}}}=\frac{4}{\mathrm{\pi }}+1$

$R_{\mathrm{e}\mathrm{q}}=\frac{\mathrm{\pi }}{\left(\mathrm{\pi }+4\right)}$