question_answer

Figure shows a circuit with three ideal batteries in it. The circuit elements have the following values                 \[\Delta {{V}_{{{B}_{1}}}}=3.0\,V\],         \[\Delta {{V}_{{{B}_{2}}}}=6.0\,V\] \[{{R}_{1}}=2.0\,\,\Omega ,\,\,{{R}_{2}}=4.0\,\,\Omega \] The currents \[{{i}_{1}},{{i}_{2}}\] and \[{{i}_{3}}\] as shown in the circuit have the values

A)  0.50 A, - 0.25 A, + 0.25 A

B)  0.25 A, - 0.50 A, - 0.25 A

C)  0.50 A, 0.50 A, 1,0 A

D)  - 0.25 A, - 0.50 A, 0.25 A

Correct Answer: A

Solution :

                 Given circuit In ABCD, From Kirchhoffs law                 \[2{{i}_{1}}+2{{i}_{1}}-4{{i}_{2}}+3-6=0\]                               \[4{{i}_{1}}-4{{i}_{2}}=3\]                                             ... (i) and in CDEF From Kirchhoffs law                 \[2{{i}_{3}}+4{{i}_{2}}+6-6+2{{i}_{3}}=0\]                                 \[4{{i}_{2}}+4{{i}_{3}}=0\]                                                 \[{{i}_{2}}=-{{i}_{3}}\]                    ?. (ii)                                                 \[{{i}_{1}}+{{i}_{2}}={{i}_{3}}\]   ... (iii) From Eq. (ii), we get                 \[{{i}_{1}}+{{i}_{2}}=-{{i}_{2}}\] From Eq. (i),                 \[{{i}_{1}}=-2{{i}_{2}}\]                 \[3=4(-2{{i}_{2}})-4{{i}_{i}}\]                 \[3=-12{{i}_{2}}\]                 \[{{i}_{2}}=-0.25\,A\]                 \[{{i}_{1}}=-2{{i}_{2}}=0.50\,A\] and        \[{{i}_{3}}=-{{i}_{2}}=0.25\,A\]