When a cell is connected in the secondary circuit of a potentiometer- the balance length is found to be 810 cm- When the cell is shunted with a 5 - resistor- the balance length is found to be reduced to 675 cm- The internal resistance of the cell isA- 0-5 -B- 1 -C- 1-5 -D- 2 -

The correct option is B 1 Ω Internal resistance of the cell, r = R [ I_1 I_2/l_2 ] Where l1=810 cm Balancing length when no current flows through ‘R’ l2 =67 ...

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Byju's Answer

Standard XII

Physics

Meter Bridge

When a cell i...

Question

When a cell is connected in the secondary circuit of a potentiometer, the balance length is found to be 810 cm. When the cell is shunted with a 5 $Ω$ resistor, the balance length is found to be reduced to 675 cm. The internal resistance of the cell is

0.5 Ω

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1 Ω

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1.5 Ω

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2 Ω

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Solution

The correct option is B
1 Ω

Internal resistance of the cell, $r = R [\frac{I_{1} - I_{2}}{l_{2}}]$

Where l₁=810 cm Balancing length when no current flows through ‘R’

l₂ =675 cm Balancing length when current flows through ‘R’

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When a cell is connected in the secondary circuit of a potentiometer, the balance length is found to be 810 cm. When the cell is shunted with a 5 Ω resistor, the balance length is found to be reduced to 675 cm. The internal resistance of the cell is

The correct option is B 1 Ω Internal resistance of the cell, r=R[I1−I2l2] Where l1=810 cm Balancing length when no current flows through ‘R’ l2 =675 cm Balancing length when current flows through ‘R’

When a cell is connected in the secondary circuit of a potentiometer, the balance length is found to be 810 cm. When the cell is shunted with a 5 $Ω$ resistor, the balance length is found to be reduced to 675 cm. The internal resistance of the cell is

The correct option is B
1 Ω

Internal resistance of the cell, $r = R [\frac{I_{1} - I_{2}}{l_{2}}]$

Where l₁=810 cm Balancing length when no current flows through ‘R’

l₂ =675 cm Balancing length when current flows through ‘R’