With suitable circuit diagram, show how emfs of two cells can be compared using a potentiometer.

With the help of a circuit diagram describe the method to find the internal resistance of a cell using potentiometer.

A potentiometer wire of length 100 cm has resis tance of 10 ohms. It is connected in series with a resistance R and a cell of emf 2V and negligible in ternal resistance. The circuit is as shown below. - What is the principle of a potentiometer ?

How can you increase the sensitivity of a potentiometer?

You are supplied with a 1m long uniform resistance wire of resistance 3Ω and a cell of emf 1.5v.Draw the connection diagram to compare the emf of Leclanche cell and Daniel cell using this arrangement.

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire, has an emf of 2V and negligible internal resistance. The potentiometer itself is 4m long. When the resistance 'R_(1)' , connected across the given cell has values of (i) infinity (ii) 9.5Omega , the balancing length on the potentiometer wire are found to be 3m and 2.85m respectively. The value of internal resistance of the cell is

Draw the circuit diagram where a zener diode is used as a direct voltage regulator