Figure shows a potentiometer with a cell of 2.0 V and internal resistance `0.40 Omega` 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 mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current drawn from the standard cell,a very high resistance of `600 k Omega` 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 is the value of E ?

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega 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 mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current drawn from the standard cell,a very high resistance of 600 k Omega 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 600 k Omega have?

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega 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 mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current drawn from the standard cell,a very high resistance of 600 k Omega 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 the internal resistance of the driver cell ?

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega 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 mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current drawn from the standard cell,a very high resistance of 600 k Omega 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 . Would the circuit work well for determining an extremely small emf, say of the order of a few m V (such as the typical emf of a thermo-couple) ? If not , how will you modify the circuit ?

A voltmeter has a resistance of 100 Omega . What will be its reading when it is connected across a cell of emf 2 V and internal resistance 20 Omega ?

A cell of emf 1.1 V and internal resistance r is connected across as external resister of resistance 10r. Find the potential difference across the external resistor.

A battery of emf 12 V and internal resistance 3 Omega is connected to a resistor. If the current in the circuit is 0.5 A, Calculate resistance of resistor.

Choose correct alternative. emf of a cell is 2.2 volt. When resistance R = 5 Omega is connected in series potential drop across the cell becomes 1.8 volt. Value of internal resistance of the cell is

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell.The balance point of the cell in open circuit is 76.3 cm.When a resistor of 9.5 Omega is,used in the external circuit of the cell,the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.