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B
2A
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C
25A
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D
156A
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Solution
The correct option is D156A Applying KVL in upper loop, −3+2I1+10I=0 ⇒I1=3−10I2......(i) Similarly, in lower loop, −10I−5+3(I1−I)=0 ⇒−13I+3I1−5=0 ⇒−13I+3(3−10I2)−5=0 ⇒−26I+9−30I−10=0 ⇒I=−156A (-ve sign indicates the current direction is opposite to what we assumed) ∴|I|=156A