A cell balances against a length of 200 cm on potentiometer wire when it is shunted by a resistance of 20 . The balancing length reduces by 25 cm, when it is shunted by 15 . The internal resistance of the cell will be?
Answers
The internal resistance of the cell is r = 2.667 Ω
Explanation:
R2 = 4Ω, l'2 = 160 cm
To find: i = ?
Internal resistance of cell r
r = R (l1 - l2 / l2)
r = 8 (l1 - 200 / 200 )
For second condition:
r = 4 (l1 - 160 / 160 )
From equation 1 and 2 "l1" can be calculated.
8 (l1 - 200 / 200 ) = 4 (l1 - 160 / 160 )
l1 - 200 / 25 = l1 - 160 / 40
25 l1 - 4000 = 40 l1 - 8000
15 l1 = 4000
l1 = 266.67 cm
Now
r = 8 (l1 - 200 / 200 ) = 8 ( 266.67 - 200 / 200 )
r = 2.667 Ω
Thus the internal resistance of the cell is r = 2.667 Ω
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The internal resistance of the cell will be 15 Ω.
Explanation:
Let the balancing length be l₁
Let the internal resistance be r
r = R (l₁ - l₂/l₂)
When the shunted resistance is 20 Ω
r = 20 (l₁ - 200/200) ⇒ (1)
When the shunted resistance is 15 Ω
r = 15 (l₁ - 175/175) ⇒ (2)
From equation (1) and (2), we get,
20 (l₁ - 200/200) = 15 (l₁ - 175/175)
4 (l₁ - 200/200) = 3 (l₁ - 175/175)
4 × 175 (l₁ - 200) = 3 × 200 (l₁ - 175)
4 × 35 (l₁ - 200) = 3 × 40 (l₁ - 175)
4 × 7 (l₁ - 200) = 3 × 8 (l₁ - 175)
28 (l₁ - 200) = 24 (l₁ - 175)
28l₁ - 5600 = 24l₁ - 4200
28l₁ - 24l₁ = 5600 - 4200
4l₁ = 1400
l₁ = 350 cm
Now, the internal resistance is:
r = 20 (350 - 200/200)
r = 20 (0.75)
∴ r = 15 Ω