Question 3.23: Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?
Class 12 - Physics - Current Electricity Current Electricity Page-131
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For comparison of two resistances, R and X resistors should be joined in series and same current should flow through them.
from Ohm's law, V= IR = kl₁
here, R = 10 ohm, l₁ = 58.3 cm
so, I(10) = k × 58.3 ------(1)
similarly, IX = kl₂
here, l₂ 68.5 cm so, IX = k × 68.5 ----(2)
dividing equations (2) ÷ (1),
IX/I(10) = k × 68.5/k × 58.3
X/10 = 68.5/58.3
X = 685/58.3 = 11.75 Ω
if we fail to find balance point with the cell of emf ε , in that case we should reduced the current I and for that emf ε of the cell responsible for current I should reduce or a resistor can also be attached in series with R and X to reduce current.
from Ohm's law, V= IR = kl₁
here, R = 10 ohm, l₁ = 58.3 cm
so, I(10) = k × 58.3 ------(1)
similarly, IX = kl₂
here, l₂ 68.5 cm so, IX = k × 68.5 ----(2)
dividing equations (2) ÷ (1),
IX/I(10) = k × 68.5/k × 58.3
X/10 = 68.5/58.3
X = 685/58.3 = 11.75 Ω
if we fail to find balance point with the cell of emf ε , in that case we should reduced the current I and for that emf ε of the cell responsible for current I should reduce or a resistor can also be attached in series with R and X to reduce current.
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