A cell of emf (E) and internal resistance (r) is balanced across (l) length of potentiometer wire. If another cell of emf 2E and internal resistance (2r) is connected in parallel to the first cell , then the balancing length will be

In an experiment on measurements of emf of a cell by a potentiometer , the balancing length for a cell of emf E and internal resistance r is found to be I.Now if another cell of emf E and internal resistance 2r is connected in parallel to the first cell and balancing length determined , then the balancing length will be: "],[" 1) "(l)/(2)],[" 2) "21],[" 3) "(1)/(2)l],[" 4) "1]

A cell of emf E and internal resistance r connected in the secondary gets balanced against length l of potentiometer wire. If a resistance R is connected in parallel with the cell, then the new balancing length for the cell will be

A cell of emf 'E' and internal resistance r connected in the secondary gets balanced against length l of potentiometer wire. If a resistance 'R' is connected in parallel with the cell, then the new balancing length for the cell will be

If n cells each of emf E and internal resistance rare connected in parallel, then the total emf and internal resistances will be

If n cells each of emf E and internal resistance rare connected in parallel, then the total emf and internal resistances will be

A potentiometer arrangement is shows in Fig. 6.62 . The driver cell has emf e and internal resistance r . The resistance of potentiometer wire AB is R.F is the cell of emf e//3 and internal resistance r//3 . Balance point (J) can be obtained for all finite value of

A potentiometer arrangement is shows in Fig. 6.62 . The driver cell has emf e and internal resistance r . The resistance of potentiometer wire AB is R.F is the cell of emf e//3 and internal resistance r//3 . Balance point (J) can be obtained for all finite value of

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 -

is made of uniform material and cross-sectional area, and it has uniform resistance per unit length. The potential gradient depends upon the current in the wire. A potentiometer with a cell of emf 2 V and internal resistance 0.4 Omega is used across the wire AB . A standard cadmium cell of emf 1.02 V gives a balance point at 66 cm length of wire. The standard cell is then replaced by a cell of unknows emf e (internal resistance r ), and the balance. Point found similarly turns out to be 88 cm length of the wire. The length of potentiometer wire AB is 1 m . If the resistance is connected across the cell e , the balancing length will

n cells, each having emf. E and internal resistance r are connected in parallel. What is the net internal resistance of the combination?