In a potentiometer circuit, when two cells of e.m.f. $1.5 \text{ V}$ and $1.2 \text{ V}$ are connected to assist each other, the balancing length is $270 \text{ cm}$. What will be the balancing length in $\text{cm}$ when these two cells are connected in opposition?

  • A
    $90$
  • B
    $81$
  • C
    $60$
  • D
    $30$

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Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

The 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 up to a few $mA$) gives a balance point at $67.3 \; cm$ length of the wire. To ensure very low currents are 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 $emf$ $\varepsilon$ and the balance point found similarly,turns out to be at $82.3 \; cm$ length of the wire.
$(a)$ What is the value of $\varepsilon ?$
$(b)$ What purpose does the high resistance of $600 \; k \Omega$ have?
$(c)$ Is the balance point affected by this high resistance?
$(d)$ Would the method work in the above situation if the driver cell of the potentiometer had an $emf$ of $1.0 \; V$ instead of $2.0 \; V ?$
$(e)$ 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 thermocouple)? If not,how will you modify the circuit?

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