• # question_answer An ac source has an internal resistance of${{10}^{4}}\Omega .$ The turn ratio of a transformer so as to match the source to a load of resistance $10\Omega$ is A) $4.62\times {{10}^{-2}}$                             B) $2.03\times {{10}^{-2}}$ C) $3.16\times {{10}^{-2}}$                             D) $5.62\times {{10}^{-2}}$

For primary & secondary circuit, ${{V}_{s}}{{l}_{s}}={{V}_{p}}{{l}_{p}}$where symbols have their usual meaning. $\Rightarrow$$\frac{V_{s}^{2}}{{{Z}_{5}}}=\frac{V_{p}^{2}}{{{Z}_{p}}}$$\Rightarrow$$\frac{V_{s}^{2}}{{{Z}_{5}}}=\sqrt{\frac{Z_{s}^{{}}}{{{Z}_{p}}}}=\sqrt{\frac{10}{{{10}^{4}}}}=\frac{\sqrt{10}}{100}$ So turns ratio, $\frac{V_{s}^{{}}}{{{V}_{p}}}=$turn ratio $=\frac{\sqrt{10}}{100}=3.16\times {{10}^{-2}}$