A cell of e.m.f 12v is balanced by 150 cm of potentiometer wire when t...
Calculation of Internal Resistance of a Cell
Given Information:
- EMF of the cell = 12V
- Length of potentiometer wire for balancing the cell = 150cm
- Resistance of shunt = 4ohm
- Length of potentiometer wire after reducing balancing length = 120cm
Solution:
Let the internal resistance of the cell be 'r'.
For balancing the cell, we have:
V = I(R+r) ... (1)
Where, V is the EMF of the cell, I is the current through the circuit, R is the resistance of the potentiometer wire, and r is the internal resistance of the cell.
At the balancing length, the potential difference across the shunt is equal to the potential difference across the potentiometer wire.
Therefore, we have:
Ir = V - I(R+x) ... (2)
Where, x is the balancing length.
From equations (1) and (2), we get:
Ir = V - IR - Ir
Or, I(R+r) = V - I(R+x)
Or, IR + Ir = V - Ix
Or, I = V / (R+r)
Substituting the value of I in equation (2), we get:
r = (x / (150 - x)) * R
Substituting the given values, we get:
r = (30 / 120) * 4 = 1 ohm
Answer:
Therefore, the internal resistance of the cell is 1 ohm.