Question

Figure shows a potentiometer with a cell of emf $2.0 V$ and internal resistance $0.04 Ω$ maintaining a potential drop across the potentiometer wire $AB .$ A standard cell which maintains a constant emf of $1.02 V$ (for very moderate currents up to a few ampere) gives a balance point of $67.3 cm$ length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600 kΩ$ 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.

$(a)$ What is the value of $E$ ?

$(b)$ What purpose does the high resistance of $600 kΩ$ have ?

$(c)$ Is the balance point affected by this high resistance?

$(d)$ Is the balance point affected by the internal resistance of the driver cell?

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

$(f)$ Would the circuit work well for determining extremely small emf, say, of the order of few mV (such typical emf of thermocouple)?

Accepted Answer

A constant emf of the given standard cell $E_{1} = 1 .02 V$ , the balance point on the wire, $l_{1} = 67 .3 c m$ , now, let the standard cell is replaced by a cell of unknown emf $67.3 cm$ , so the new balance point on the wire $l_{2} = 82.3 cm$ .
(a) The location of the balance point on the wire is proportional to the emf, i.e.,

$\frac{E_{1}}{E} = \frac{l_{1}}{l_{2}}$
$\Rightarrow E = \frac{l_{2}}{l_{1}} \times E_{1}$ $= \frac{82 .3}{67 .3} \times 1 .02 = 1 .247 V$
Hence, the value of the unknown emf is $1.247 V$ .

(b) The high resistance of $600 kΩ$ is used to lower the current flow through the galvanometer when the movable contact is far from the balance point. It decreases the heat dissipated, so it ensures that the value of resistances does not change with time.
(c) When the balance point is found, there will not be any current or voltage across the $600 kΩ$ Resistor. Therefore, it is as good as a simple wire. Hence, it will have no effect on the balance point.

(d) The internal resistance of the driver cell does not affect the balance point.

(e) If the potentiometer's driver cell emf is less than the other cell's emf, then there would be no balance point on the wire. Hence, the method won't work if the potentiometer's driver cell has an emf of $1 V$ instead of $2 V$ .

(f) The circuit cannot work well for determining a tiny emf as there will be a very high percentage of error. This is because the circuit would be unstable; the balance point would be very close to end $A$ . A series resistance can be connected to the wire $A B$ at the end $B$ , such that the potential drop in $A B$ is slightly above the small emf to be found. This way, the balance point can be anywhere from $A$ to $B$ . Hence, the error will be lesser.

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