12 Physics Experiment: Internal Resistance of a Cell Potentiometer

 

Potentiometer Experiment:

 Determining Internal Resistance of a Cell 

Here is the complete write-up, including a sample set of readings and the corresponding calculations for determining the internal resistance (r) of a cell using a potentiometer.

💡 Formula Used

The internal resistance r is calculated using the formula:

Where:

• R is the known external (shunt) resistance taken from the resistance box .

• L1 is the balancing length for the cell's EMF (E) (when key K2 is open) (in cm)

• L2 is the balancing length for the cell's terminal potential difference (V) (when key K2 is closed and resistance R is shunted) (in cm).

• 
  

📋 Observations and Table

S.No.	Shunt Resistance, R (Ω)	Bal. Length l1​ (cm) (for E) (Key K2​ open)	Bal. Length l2​ (cm) (for V) (Key K2​ closed)	l1/l2	Internal Resistance r=R[(l1/l2)-1] (Ω)
1	5.0	250.0	220.0	1.136	r1 = 5.0 (1.136 - 1) = 0.618 Ω
2	10.0	250.0	235.0	1.063	r2 = 10.0 (1.063 - 1)       =  0.638 Ω
3	15.0	250.0	2240.0	1.041	r3 = 15.0 (1.041 - 1) = 0.615Ω

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Note: The balancing length L1for the EMF (E) should ideally remain constant, as the EMF of the cell does not change. Therefore, it is typically measured once at the beginning, or taken as the mean of theL1 readings, and used for all calculations.

  Calculations

1. For the first set of readings (S.No. 1):

• R = 5.0 Ω 

• L1 = 250.0 cm

• L2 = 220.0 cm

• r =R[(l1/l2)-1]

• 5.0 [(250/220)-1]

• 5 [1.136 - 1]

• 5[0.136]

• r = 0.618 Ω

2.    R = 10.0Ω

• L1 = 250.0 cm

• L2 = 235.0 cm

• r =R[(l1/l2)-1]

      r = 10 [(250/235)-1]

         =10 [1.063 - 1]

          = 10 [0.063]

          = 0.638 Ω

     

• R = 15.0Ω

• L1= 250.0 cm

• L2 = 240.0 cm

• r = 15 [(250/240)-1]

•   = 15 [ 1.041 - 1 ]

•   = 15[0.041]

•    = 0.615 Ω

📊 Result

Mean Internal Resistance r =[ (0.618 +  0.638 + 0.615) /3]

= (1.871)/3

Mean Internal Resistance r = 0.623 Ω

Result: The internal resistance of the given primary cell is  r = 0.623 Ω

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