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 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 a potentiometer circuit, the emf of driver cell is 2V and internal resistance is 0.5 Omega . The potentiometer wire is 1m long . It is found that a cell of emf 1V and internal resistance 0.5 Omega is balanced against 60 cm length of the wire. Calculate the resistance of potentiometer wire.

In a connection of 10 dry cells each of emf E and internal resistance r in series, two cell is connected opposite. Net emf and net internal resistance will be:

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

The potentiometer wire of resistance R is connected in series with a cell of e.m.f. E and of resistance R_(h) . The current flowing through the potentiometer wire is

The e.m.f. of a cell is E volts and internal resistance is r ohm. The resistance in exteral circuit is also r ohm. The p.d across the cell will be

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 network as shown in the figure the potential difference across the resistance 2R is (the cell has an emf of E and has no internal resistance):

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

What is the maximum power output than can be obtained from a cell of emf E and internal resistance r?