Very high or very low resistance cannot be measured correctly by using the Wheatstone bridge principle. Give reason.

A very low resistance cannot be measured with a wheatstone bridge . Why ?

How resistance can be determined using the principle of Wheatstone bridge ? Derive the principle .

A coil of wire is kept in melting ice . Its resistance measured by a Wheatstone bridge is 5Omega . If the coil is heated to 100^(@)C and another wire of resistance 100Omega is coonnected in parallel to it , the balanced condition of the brige remains unchaged . Determine the temperature coefficient of resistance of the wire of the coil .

The value of an unknown resistance is found to be 8Omega when it is determined with the help of a wheatstone bridge by using known reistance 100Omega , 10Omega and 80 Omega . The emf of the electric cell used is 2 v and its internal resistance is 1.1Omega Determine the value of current flowing through the cell .

A student chooses the stancdard resistance S to be 100Omega while measuring a resistance R by using a metre bridge He finds the null point at l_(1)=2.9 cm . He is told to attempt to improve accuracy . Which of the following is a useful way ?

Statement I : If the resistance of the first two arms P and Q of a balanced Wheatstone bridge are exchanged , the balanced condition is not disturbed . Statement II : The balanced condition of a Wheatstone bridge is independent of the resistance of the galvanometer used .

Explain from vander Waal's equation that at very low pressure and at very high temperature the real gases behave as an ideal gases.

Fig. shows a potentiometer with a cell of 2.0V and internal resistance 0.40Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02V (for very moderate currents upto a few pA) gives a balance point at 67.3 cm length of the wire. Io ensure very low currents drawn from the standard cell, a very high resistance of 600kOmega is put in series with it which is shorted close to the balance point. The standard cell is then, replaced by a cell of unknown emf e and the balance point found similarly turns out to be at 82.3 cm length of the wire, What purpose does the high resistance of 600kOmega have?

Fig. shows a potentiometer with a cell of 2.0V and internal resistance 0.40Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02V (for very moderate currents upto a few pA) gives a balance point at 67.3 cm length of the wire. to ensure very low currents drawn from the standard cell, a very high resistance of 600kOmega is put in series with it which is shorted close to the balance point. The standard cell is then, replaced by a cell of unknown emf e and the balance point found similarly turns out to be at 82.3 cm length of the wire, Is the balance point affected by this high resistance?

Two identical cells of emf 1.5V each joined in parallel supply energy to an external circuit consisting of two resistance of 7 Omega each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4V.Calculate the internal resistance of each cell.