in a meter bridge when the resistance in the left gap is 2ohm & an unknown resistance in the right gap, the balance point is obtained at 40cm from zero end. on shunting the unknown resistance with 2ohm, find the shifty of the balance point on the bridge wire.
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Answered by
14
Initially resistances are R and S in the left and right gap of meter bridge.
According to formula , R/S=l/100-l.
L is given 40
Therefore R/S = 2/3
s=3R/2.
After connecting resistance 10 in series with R
R+10/S=3/2 (given that l=60)
Substituting s=3R/2 we get
2R+20/3R = 3/2
5R =40 thus R=8.
s=12.
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According to formula , R/S=l/100-l.
L is given 40
Therefore R/S = 2/3
s=3R/2.
After connecting resistance 10 in series with R
R+10/S=3/2 (given that l=60)
Substituting s=3R/2 we get
2R+20/3R = 3/2
5R =40 thus R=8.
s=12.
hope this helps you
mark it the brainliest
Answered by
5
Answer:
Explanation:22.5cm
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