Figure 3.33 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 1.02 V (for very moderate currents upto a few mA) 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 `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 epsilon` 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? :

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 1.02 V (for very moderate currents upto a few mA) 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 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 epsilon and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What is the value epsilon ? :

Figure 3.33 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 1.02 V (for very moderate currents upto a few mA) 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 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 epsilon 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 1.02 V (for very moderate currents upto a few mA) 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 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 epsilon 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 kOmega have? :

Figure 3.33 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 1.02 V (for very moderate currents upto a few mA) 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 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 epsilon 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 mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?:

Find the current drawn from a cell of emf 1V and internal resistance 2/3 Omega connected to the network given adjoint.

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A current of 2.0 A passes through a cell of emf 1.5 V having internal resistance of 0.15 Omega . The potential difference (measured in volt) across both the terminals of the cell is

A cell of emf 2 V and internal resistance 0.1Omega is connected to 3.9Omega external resistance. What will be the potential difference across the terminals of the cell?