CEE Kerala Engineering
CEE Kerala Engineering Solved Paper-2005
question_answer
The currents ii and 13 through the resistors\[{{R}_{1}}(=10\,\Omega )\]and\[{{R}_{2}}(=30\,\Omega )\]in the circuit -diagram with\[{{E}_{1}}=3V\,,{{E}_{2}}=3V\]and\[{{E}_{3}}=2V\]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
E) 0.4 A, 0.1 A
Correct Answer:
A
Solution :
In closed loop EFGDE \[{{i}_{2}}{{R}_{2}}={{E}_{2}}\] \[{{i}_{2}}\times 30=3\] \[{{i}_{2}}=0.1\,A\] In closed loop ABCEA \[-{{i}_{1}}{{R}_{1}}-{{E}_{1}}+{{E}_{2}}+{{E}_{3}}=0\] \[-{{i}_{1}}\times 10-3+3+2=0\] \[{{i}_{1}}=0.2A\]