In a potentiometer experiment- the balancing with a cell is at 240cm-On shunting the cell with a resistance of 2 Omega- the balancing length becomes 120cm-Find the internal resistance of the cell-

Here balancing length (L1) = 240 cm, when shunted with 2 nbsp; Omega; resistance new balancing length become = 120 cm r'=l1 l2l2 #215;r=240 120120 #215; ...

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Standard XII

Physics

V-I Characteristics for a P-N Junction Diode

In a potentio...

Question

In a potentiometer experiment, the balancing with a cell is at 240cm.On shunting the cell with a resistance of 2 Ω, the balancing length becomes 120cm.Find the internal resistance of the cell.

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Solution

Here balancing length (L₁) = 240 cm, when shunted with 2 Ω resistance new balancing length become = 120 cm
$r' = (\frac{l_{1} - l_{2}}{l_{2}}) \times r = (\frac{240 - 120}{120}) \times 2 = 2 Ω$
where r' = internal resistance

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In a potentiometer experiment, the balancing with a cell is at 240cm.On shunting the cell with a resistance of 2 Ω, the balancing length becomes 120cm.Find the internal resistance of the cell.

Here balancing length (L1) = 240 cm, when shunted with 2 Ω resistance new balancing length become = 120 cm r'=l1-l2l2×r=240-120120×2=2Ω where r' = internal resistance

Here balancing length (L₁) = 240 cm, when shunted with 2 Ω resistance new balancing length become = 120 cm
$r' = (\frac{l_{1} - l_{2}}{l_{2}}) \times r = (\frac{240 - 120}{120}) \times 2 = 2 Ω$
where r' = internal resistance