the resistances in the two arms of metre Bridge are 5Ω and R Ω respectively when the resistance R is shunted with an equal resistance the new balance point is at 1.6 l1 the resistance R, is
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=> R= 15 Ω
For the first case balance condition if meter bridge will be
5/l1 = R/ ( 100-l1 ) _[i]
Now, by shunting resistance R by an equal resistanceR, new resistance and that am become R/2 So, new balance condition will be
5/1.6 l1 = R/2 / ( 100 - 1.6 l1 ) _[ii]
From Eqs [i] and [ii],
1.6 /1 = [ ( 100 - 1.6 l1 ) / 100-l1 ] × 2
=> 160 - 1.6 l1 = 200 - 3.2 l1
1.6 l1 = 40
l1 = 40/1.6 = 25 m
=> from Eq [i], 5/25 = R/75
=> R = 15Ω
∴The resistance R is 15Ω !
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Answer:
♣️The resistance R is 15 Ω♣️
♥️Hope it will be helpful ☺️ thank you!♥️
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