Home
Class 12
PHYSICS
In a potentimeter arrangment for determi...

In a potentimeter arrangment for determining the emf of a cell , the balance point of the cel in open circuit is 350 cm . When a resistance of `9Omega` is used in the external circuit of h the balance point shift to 300cm . Determine the internal reistance of the cell .

Text Solution

Verified by Experts

The correct Answer is:
2
Here , `l_(1)=350:cm",l_(2)=300"cm" .R=9Omega`
Internal reistance of the cell ,
`r=((l_(1))/(l_(2))-1)R=((350)/(300)-1)xx9=1.5Omega`
Promotional Banner

Topper's Solved these Questions

  • KIRCHHOFF'S LAWS AND ELECTRICAL MEASUREMENT

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE (NEET)|1 Videos

  • ELEMENTARY PHENOMENA OF ELECTROSTATICS

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE|2 Videos

  • LIGHT WAVE AND INTERFERENCE OF LIGHT

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|15 Videos

Similar Questions

Explore conceptually related problems

Fig. shows a 2.0V potentiometer used for the determination of internal resistance of 1.5V cell- The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Omega is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

The electromotive force of a cell is 2V. The potential difference becomes 1.5 V when a resistance of 15 Omega is added to the two ends of the cell. Determine the internal resistance of the cell and the lost volt.

In a potentiometer experiment , a particular cell produces balance in the wire at 560 cm . If a 10Omega resistance is connected with the cell in parallel ,then this balanced lendth change to 412 cm . Find out the internal resistance of the cell .

In a metre-bridge [Fig.], the null point is found to be at a distance of 33.7 cm from A. If now a resistance of 12Omega is connected in parallel with S, the null point occurs at 51.9 cm. Determine the resistance of R and S.

The potential difference between the terminals of a cell, in an open circuit was 2.5 volt and when it was connected through a resistance of 5 ohm, the potential difference fell to 1.5v. Find the internal resistance of the cell.

A length of potentiometer wire of 155 cm balances the emf of a cell in a circuit and a length of 135 cm when the cell has a conductor of resistance 8Omega connected between its terminals . Find the internal resistance of the cell .

In a potentiometer experiment . A cell produces balance in the wire at 240 cm. balancing length change to 120 cm when a resistance 2Omega Is connected parallel to the cell. Find out the internal resistance of the cell.

In a potentimeter experiment , it is found that no current passes through the galvanometer when the termials of the cell are connected across 52 cm of the potentimeter wire . If the cell is shunted by a resistance of 5Omega , a balance is found when the cell is connected across 40 cm of the wire . Find the internal resistance of ht cell .

5 Identical cells are connected in series to form a row, 3 such rows are connected in parallel to form a combination of cells. The combination sends 1A current through an external resistance of 9 Omega . But if the resistance of the external circuit be 19 Omega then the currents is reduced to half of its previous value. Determine the em and internal resistance of each coil.