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?

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, Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

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?

A cell of emf 2Vand internal resistance 1.5Omega is connected to two ends of 100 cm wire. The resistance of the wire is 0.005 Omega//cm. The potential gradient of the wire is

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When a resistance of 12 Omega is conencted with a cell of emf 1.5 V,0.1 A current flows through the resistance internal resistance of the cell is

When a 50Omega wire is connected extremely with 11 volt emf cell having internal resistance 50 Omega then find potential difference and electric current.

What is the minimum number of cells each of emf 10 V and internal resistance 1 Omega to pass a current of 10 A through a resistance R of 3 Omega ? What is the discharging power of R?

Fig. shows a 2.0V potentiometer used for the determination of internal resistance of 1.5V cell- The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Omega 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.

When a resistance of 12 Omega is connected with a cell of emf 1.5 V, 0.1A current flows through the resistance. Internal resistance of the cell is—