In order to measure internal resistance r1 of a cell emf E, a meter bridge of wire resistance `R_0 = 50 Omega`, a resistance `R_0/2` another cell of emf E/2 and galvanometer G are used in circuit. If null point is founded at l = 72cm. then value of `r1`

Define internal resistance of a cell. Prove that r ~= ((E)/(V) - 1)R , where R is the external resistance used.

Internal resistance of battery of emf 2E is R_1 and battery of emf E is R_2 . If the potential difference across the cell of emf 2E is zero, then value of R is

A potentiometer wire of length 100 cm has a resistance of 10 Omega . It is connected in series with a resistance R and cell of emf 2 V and of of 40 cm of the potentiometer wire. What is the value of R ?

Null point in the galvanometer is obtained when a cell of emf E and internal resisance r is connncted across the length of 22 cm wire of the potentiometer . Now a resistane of 10Omega is connected across the terminals of the cell (by closing the key K) and null point is obtained against the length of cm. The internal resistance r of the cell is -

The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

A cell of emf E having an internal resistance R varies with R as shown in figure by the curve

If E is the emf of a cell of internal resistance r and external resistance R, then potential difference across R is given as

In a potentiometer experiment of measuring the e.m.f. of a cell, the null point is at 300 cm, when we have a resistance of 200 Omega in series with the cell and the galvanometer. If the resistance is increased to 300 Omega , the null point will be shifted through