• # question_answer In the circuit the figure. When key ${{K}_{1}}$is closed, the ammeter reads${{I}_{0}}$whether${{K}_{2}}$is open or closed. But, when${{K}_{1}}$is open, the ammeter reads${{I}_{0}}/2,$when${{K}_{2}}$is closed. Assuming that ammeter resistance is much less than ${{R}_{2}},$find the values of rand${{R}_{1}}$(in ohms) A)  100, 50                       B)  50, 100C)  0, 100             D)  0, 50

When${{K}_{1}}$is closed, the resistance${{R}_{1}}$is ineffective, Applying kirchoff's voltage law with${{k}_{1}}$closed and${{K}_{2}}$open, while moving through ${{k}_{1}},{{k}_{2}}$and the battery, we get ${{I}_{0}}{{R}_{2}}+{{I}_{0}}r-\varepsilon =0\Rightarrow {{I}_{0}}=\frac{\varepsilon }{{{R}_{2}}+r}$ Applying Kirchoff?s voltage law with both ${{k}_{1}}$and${{k}_{2}}$closed, while moving through ${{k}_{1}},{{k}_{2}}$ and the battery, we get ${{I}_{0}}{{R}_{2}}+(2{{I}_{0}})r-\varepsilon =0\Rightarrow {{I}_{0}}=\frac{\varepsilon }{{{R}_{2}}+2r}$ Equating the values of ${{I}_{0}},$we get $\frac{\varepsilon }{{{R}_{2}}+r}=\frac{\varepsilon }{{{R}_{2}}+2r}\Rightarrow r=0$ Applying Kirchoff's voltage law while moving through ${{R}_{1}},{{k}_{2}}$and the battery, with ${{k}_{1}}$ open and${{k}_{2}}$closed, we get ${{I}_{0}}{{R}_{1}}+\frac{{{I}_{0}}}{2}{{R}_{2}}-\varepsilon +{{I}_{0}}r=0\Rightarrow {{I}_{0}}=\frac{\varepsilon }{{{R}_{1}}+\frac{{{R}_{2}}}{2}+r}$ $\therefore$    $\frac{\varepsilon }{{{R}_{1}}+\frac{{{R}_{2}}}{2}+r}=\frac{\varepsilon }{{{R}_{2}}+r}$ $\Rightarrow$            $\frac{1}{{{R}_{1}}+50}=\frac{1}{100}\Rightarrow {{R}_{1}}=50\Omega$ Hence, the correction option is [d].