question_answer

The current \[{{i}_{1}}\] and \[{{i}_{2}}\]through the resistors \[{{R}_{1}}(=10\,\Omega )\] and \[{{R}_{2}}(=30\,\Omega )\] on the circuit diagram with \[{{E}_{1}}=3\,V\], \[{{E}_{2}}=3\,V\] and \[{{E}_{3}}=2\,V\] are respectively

A)  \[0.2\] A, \[0.1\] A         

B)  \[0.4\] A, \[0.2\] A

C)  \[0.1\] A, \[0.2\] A         

D)  \[0.2\] A, \[0.4\] A

Correct Answer: A

Solution :

In closed loop EFGDE \[{{i}_{2}}{{R}_{2}}={{E}_{2}}\] \[{{i}_{2}}\times 30\,=3\] \[{{i}_{2}}=0.1A\] In closed loop ABCEA \[I=-{{i}_{1}}{{R}_{1}}\,-{{E}_{1}}+{{E}_{2}}+{{E}_{3}}=0\] So    \[{{I}_{1}}=0.2A\]