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; ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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.

Open in App

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

Suggest Corrections

Similar questions

Q. In a potentiometer experiment, the balancing with a cell is at length

240 c m

. On shunting the cell with a resistance of

2 Ω

, the balancing length becomes

120 c m

. The internal resistance of the cell is

Q. In a potentiometer experiment, experimental cell used in secondary circuit is balanced against

240 c m

length of potentiometer wire. On shunting this experimental cell with a resistance of

2 Ω,

the balancing length becomes

120 c m .

The internal resistance of the cell is

Q. In a potentiometer the balancing with a cell is at length of

220

cm. On shunting the cell with a resistance of

3 Ω

balance length becomes

130

cm. What is the internal resistance of this cell?

Q. In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a

5 Ω

resistance; and is at a length of 3 m when the cell is shunted by a

10 Ω

resistance. The internal resistance of the cell is, then

Q. In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of

2 m

, when the cell is shunted by a

5 Ω

resistance and at a length of

3 m

, when the cell is shunted by a

10 Ω

resistance. The internal resistance of the cell is

Join BYJU'S Learning Program

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