A) 2 A, 4 A
B) 4 A, 2 A
C) I A, 2 A
D) 2 A, 3 A
Correct Answer: A
Solution :
From Kirchhoffs first law at junction P, \[{{I}_{1}}+{{I}_{2}}=6\] ...(i) From Kirchhoffs second law to the closed circuit PQRP, \[-2{{I}_{1}}-2{{I}_{1}}+2{{I}_{2}}=0\] \[\Rightarrow \] \[-4{{I}_{1}}+2{{I}_{2}}=0\] \[\Rightarrow \] \[2{{I}_{1}}-{{I}_{2}}=0\] ?(ii) Adding Eqs. (i) and (ii), we get \[3{{I}_{1}}=0\] \[\Rightarrow \] \[{{I}_{1}}=2A\] From Eq. (i), \[{{I}_{1}}=6-2=4A\]
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