A) 0.5 V
B) 0.4 V
C) 0.5 V
D) 0.3 V
Correct Answer: D
Solution :
In first case, current \[{{i}_{1}},\] flowing through secondary circuit\[{{i}_{1}}=\frac{6}{{{R}_{AB}}+2}=1A\] Now, once the balance point is achieved, \[{{\varepsilon }_{1}}={{R}_{AJ}}{{i}_{1}}\] \[\Rightarrow \]\[0.5={{R}_{AJ}}(1)\] ?.(i) Similarly, in the second case, \[{{\varepsilon }_{2}}={{R}_{AJ}}{{i}_{2}}\] \[\Rightarrow \]\[{{\varepsilon }_{2}}={{R}_{AJ}}\left( \frac{6}{4+6} \right)\Rightarrow {{\varepsilon }_{2}}=\frac{6}{10}{{R}_{AJ}}\] ?.(ii) From eqn. (i) \[{{\varepsilon }_{2}}=\frac{6}{10}\times 0.5\Rightarrow \varepsilon =0.3V\]You need to login to perform this action.
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