Question
Download Solution PDFIn the circuit shown below, the value of the current I is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct option is 2
Concept:
Applying KCL at Node B
I+5 = I3
Where I3 is the current flowing through 3 Ω resistance.
Applying KVL in the outer loop
20 - 2I -3(I3) - 3I = 0 ..................(1)
Substituting the value of I3 in eq(1) , we have:
20 - 2I - 3(5+I) -3I=0
⇒20 -2I -15-3I-3I=0
⇒8I = 5
⇒I =5/8 A
Last updated on Jul 1, 2025
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