The formula for internal resistance of a...

The formula for internal resistance of a cell

`R((l_2-l_1)/(l_2))`

`R((l_1-l_2)/(l_2))`

`R((l_1)/(l_1-l_2))`

`R((l_2)/(l_2-l_1))`

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A potentiometer circuit has been setup for finding. The internal resistance of a given cell. The main battery used a negligible internal resistance. The potentiometer wire itsefl is 4 m long. When the resistance, R , connected across the given cell, has value of (i) Infinity 9.5 Omega , (ii) the 'balancing length' , on the potentiometer wire are found to be 3 m and 2.85 m , respectively. The value of internal resistance of the cell is

Name the device used for the measurement of internal resistance of a cell.

The e.m.f. of a cell is balanced by fall of potential along 281 cm of potentiometer wire. If a resistance of 8 Omega is connected across the cell , the balancing length is found to be 151 cm. Calculate internal resistance of a cell.

When a balance point is obtained in a potentiometer for finding the internal resistance of a cell, the current through the potentiometer wire is due to

Describe with the help of neat diagram how will you determine the internal resistance of a cell by using potentiometer. Derive the necessary formula.

On what factors does the internal resistance of a cell depend?

Choose correct alternative. The internal resistance of a cell of emf 2V is 0.1 Omega . It is connected to a resistance of 0. 9 Omega . The voltage across the cell will be.

Choose correct alternative. emf of a cell is 2.2 volt. When resistance R = 5 Omega is connected in series potential drop across the cell becomes 1.8 volt. Value of internal resistance of the cell is

Choose correct alternative. A battery of emf 10 volt is connected to a voltmeter of resistance 135 Omega . Find the reading of the voltmeter if internal resistance of the cell is 15 Omega A. 8 volt B. 9 volt C. 10 volt D. none of the above

CHETANA PUBLICATION-Current Electricity-Exercise

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