question_answer

Three batteries of negligible internal resistance and three resistors of 4, 8 and 12 ohm are connected as shown in figure here. The current through the 12 ohm resistor is :

A) \[0.57A\]

B) \[1.09A\]

C) \[0.04A\]

D) \[1.14A\]

Correct Answer: D

Solution :

In close loop\[BCEDB\],

\[-4\left( i-i' \right)+10-12i+8=0\]

and in close loop\[ABDFA\],

\[-4\left( i-i' \right)+10+8\left( i \right)-6=0\]

On solving above equations, we get

\[i=1.14\operatorname{A}.\]