when a resistor of 5 ohm is connected across its terminals pottential difference is balanced by 150 cm of potentiometer wire and when a resistance of 10 ohm is connected across cell the terminal potential difference is balanced by 175 cm by same potentiometer wire. find the balancing length, when the cell is in open and the internal resistance of the cell?
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Given :
Balancing length when 5Ω resiatance is connected = 150cm
Balancing length when 10Ω resiatance is connected = 175cm
To Find :
Balancing length when the cell is open and internal resistance
Solution :
- For a potentiometer
r = [ (L-l)×R ] / l
- By substituting the value in the equation
r = [ (l-l₁)×R ] / l₁
r = [ (l-150)×5 ] / 150……(1)
r = [ (l-l₂)×R ] / l₂
r = [ (l-175)×10 ] / 175 ……(2)
- By equating both the equations
[ (l-150)×5 ] / 150 = [ (l-175)×10 ] / 175
7l - 1050 = 12l - 2100
l = 210 cm
The balacing length of the wire is 210cm
- From equation (1)
r = [ (l-150)×5 ] / 150
r = [ (210-150)×5 ] / 150
r = 2Ω
The internal resistance of the cell is 2Ω
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