Flag Electric Current> Please solve Q.2. πŸ₯°πŸ₯°πŸ₯°πŸ₯°πŸ˜πŸ₯°πŸ˜›πŸ˜‹πŸ˜£πŸ€«πŸ™οΏ½...
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Please solve Q.2.
πŸ₯°πŸ₯°πŸ₯°πŸ₯°πŸ˜πŸ₯°πŸ˜›πŸ˜‹πŸ˜£πŸ€«πŸ™πŸ€«πŸ™„πŸ€πŸ™„πŸ€πŸ€«πŸ§πŸ€«πŸ§πŸ€«πŸ§πŸ€«πŸ§πŸ€«πŸ§πŸ€πŸ§πŸ€πŸ˜•πŸ€πŸ§πŸ€«πŸ€«πŸ™„πŸ€«πŸ™„πŸ€«πŸ™„πŸ€«πŸ™„πŸ˜²πŸ€«πŸ€«πŸ™„πŸ€«πŸ˜ŸπŸ€πŸ˜ŸπŸ€πŸ˜•πŸ€πŸ˜ŸπŸ€πŸ˜•πŸ€πŸ€πŸ˜•πŸ€πŸ§πŸ€

Ankur , 6 Years ago
Grade 7
anser 2 Answers
Saurabh Koranglekar
576-1431_IMG_20191217_173213.jpg
Last Activity: 6 Years ago
Vikas TU
Dear student 
Simply understand this by  : In series voltage drop ccurs  so P = V^2/R  so power wil be reduced.
In parallel voltage wil be same so power will depend on resistance of material .
Hope this clears.
Good Luck 
Last Activity: 6 Years ago
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