question_answer

Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are \[\frac{1}{{{R}^{2}}}\], and \[{{R}^{2}}\]\[{{R}_{1}}\]. If the potential difference across the source having internal resistance\[{{R}_{2}}\] , is zero, then

A)  \[({{R}_{2}}>{{R}_{1}})\]             

B)  \[{{R}_{2}}\]     

C)        \[R=\frac{{{R}_{2}}\times ({{R}_{1}}+{{R}_{2}})}{({{R}_{2}}-{{R}_{1}})}\]

D)        \[R={{R}_{2}}-{{R}_{1}}\]

Correct Answer: B

Solution :

\[\omega \] \[\beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{1-0.98}=49\]  \[{{A}_{v}}=(49)\left[ \frac{500\times {{10}^{3}}}{{{R}_{1}}} \right]\] According to the question, \[=6.0625\times {{10}^{6}}=49\times \left[ \frac{500\times {{10}^{3}}}{{{R}_{1}}} \right]\times 49\] \[\therefore \] \[{{R}_{1}}=198\Omega \] \[{{S}_{x}}=(6-4)\cos {{45}^{o}}\hat{i}\] \[=2\times \frac{1}{\sqrt{2}}=\sqrt{2}km\] \[{{S}_{y}}=(6+4)\sin {{45}^{o}}\hat{j}\]   \[=10\times \frac{1}{\sqrt{2}}=5\sqrt{2}km\]