question_answer

Four cells each of emf 2.0 V and internal resistance \[1\Omega \]are connected as shown in the figure. Find \[{{I}_{1}},{{I}_{2}}\] and \[{{I}_{3}}\].

A)  0.8 A, 0.4 A, 0.4 A

B)  0.5 A, 0.25 A, 0.25 A 

C)  (8/13) A, (4/13) A, (4/13) A

D)  None of the above

Correct Answer: B

Solution :

Write KVL for CDFE \[{{I}_{2}}(2+2+1+1)-4-{{I}_{3}}(2+2+1+1)+4=0\] \[\Rightarrow \]            \[6{{I}_{2}}=6{{I}_{3}}\] \[\Rightarrow \]            \[{{I}_{2}}={{I}_{3}}\]                                          ?(i) And      \[{{I}_{2}}+{{I}_{3}}+{{I}_{1}}\]                                  ?(ii) Apply KVL for EFAB \[{{E}_{3}}(2+2+1+1)-4+{{I}_{1}}(5)=0\]                    ?(iii) After solving the Eqs. (i), (ii) and (iii), we get \[{{I}_{1}}=0.5A,{{I}_{2}}=0.25A,{{I}_{3}}=0.25A\]