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 \[{{i}_{3}}=-{{i}_{2}}=0.25\,A\]