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Grade: 7
        
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€
Β  Β  Β  Β  Β Β 
2 months ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
3441 Points
							576-1431_IMG_20191217_173213.jpg
						
2 months ago
Vikas TU
10070 Points
							
Dear studentΒ 
Simply understand this byΒ  : In series voltage drop ccursΒ  so P = V^2/RΒ  so power wil be reduced.
In parallel voltage wil be same so power will depend on resistance of material .
Hope this clears.
Good LuckΒ 
2 months ago
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