The wire in the potentiometer has a resistance of R0 and the potentiometer is connected to a battery of
voltage 'V'. Now a resistor 'R' whose value of resistance has to be measured is connected. When the
sliding point is exactly in the middle of the potentiometer, the voltage drop across 'R' is V/4. What is the
value of R/R0?
Answers
Answer:
very good question but i don't know the answer
Given : The wire in the potentiometer has a resistance of R₀ and the potentiometer is connected to a battery of voltage 'V'.
Now a resistor 'R' whose value of resistance has to be measured is connected.
When the sliding point is exactly in the middle of the potentiometer, the voltage drop across 'R' is V/4.
To Find : the value of R/R₀
Solution:
sliding point is exactly in the middle of the potentiometer
Hence potentiometer resistance = R₀ /2
a resistor 'R'
Hence total resistance 1/ Rₓ = 1/R + 1/(R₀ /2)
=> 1/Rₓ = 1/R + 2/R₀
=> 1/Rₓ = (R₀ + 2R)/RR₀
=> Rₓ = RR₀/(R₀ + 2R)
Total Rt = Rₓ + R₀ /2
=> Rt = RR₀/(R₀ + 2R) + R₀ /2
=> Rt = (2RR₀ +R₀² + 2R R₀) /2(R₀ + 2R)
=> Rt = ( R₀² + 4R R₀) /2(R₀ + 2R)
Hence current = V/ Rt
=> I = 2V(R₀ + 2R) / ( R₀² + 4R R₀)
V/4 = I Rₓ
=> V/4 = Rₓ 2V(R₀ + 2R) / ( R₀² + 4R R₀)
=> ( R₀² + 4R R₀) = 8 Rₓ (R₀ + 2R)
Rₓ = RR₀/(R₀ + 2R)
=> ( R₀² + 4R R₀) = 8 RR₀
=> R₀ + 4R = 8R
=> R₀ = 4R
=> R/R₀ = 1/4
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