JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
The current \(i\) in the network is.....\(A\)
- A \(0 \)
- B \(0.6\)
- C \(0.3 \)
- D \(0.2 \)
Answer & Solution
Correct Answer
(C) \(0.3 \)
Step-by-step Solution
Detailed explanation
\(i=\frac{9}{(5+10+5+10)}=\frac{9}{30} A\)
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