A) 1 A
B) 4 A
C) 2 A
D) 3 A
Correct Answer: C
Solution :
Let current through XY is \[{{i}_{3}}.\]Applying Kirchhoff?s law to loop (1) and (2) \[{{i}_{1}}+0\times {{i}_{3}}-3{{i}_{2}}=0\] \[{{i}_{1}}=3{{i}_{2}}\] (i) and \[-2({{i}_{1}}-{{i}_{3}})+4({{i}_{2}}+{{i}_{3}})=0\] So, \[2{{i}_{1}}-4{{i}_{2}}=6{{i}_{3}}\] (ii) Also, \[50=1{{i}_{1}}-2(i-{{i}_{3}})\] \[\therefore \] \[3{{i}_{1}}-2{{i}_{3}}=50\] (iii) From Eqs. (i), (ii), and (iii), we get \[{{i}_{2}}=2A\] Hence, the correction option is [c].You need to login to perform this action.
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