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A cell of e.m.f 12v is balanced by 150 cm of potentiometer wire when the cell is shu ted by a resistance of 4 ohm the balancing length is reduced by 30 cm what is the internal resistance of the cell?
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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.
Community Answer
A cell of e.m.f 12v is balanced by 150 cm of potentiometer wire when t...
L1=150
L2=120
r=R[(L1/L2)-1]
Hennce, r=4×(30/150)
r=0.8ohms
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A cell of e.m.f 12v is balanced by 150 cm of potentiometer wire when the cell is shu ted by a resistance of 4 ohm the balancing length is reduced by 30 cm what is the internal resistance of the cell?
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