Question: Q. 13. Calculate the value of the unknown potential $V$ for the given potentiometer circuit. The total length $(400 \mathrm{~cm})$ of the potentiometer wire has a resistance of $10 \Omega$ and the balance point is obtained at a length of $240 \mathrm{~cm}$.
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Solution:
Ans. The current through the potentiometer wire
$$ =\frac{3 \mathrm{~V}}{(290+10) \Omega}=10^{-2} \mathrm{~A} \quad \mathbf{1} $$
$\therefore$ Potential drop per unit length of the potentiometer
$$ \text { wire, } \quad \begin{aligned} \phi & =\frac{10^{-2} \mathrm{~A} \times 10 \Omega}{400 \mathrm{~cm}} \ & =\frac{1}{4} \times 10^{-3} \mathrm{~V} / \mathrm{cm} \end{aligned} $$
$$ \begin{align*} \mathrm{V} & =\phi l=\frac{1}{4} \times 10^{-3} \times 240 \text { volt } \ & =6 \mathrm{~V} \times 10^{-2} \mathrm{~V} \ & =60 \text { millivolt } \ & =60 \mathrm{mV} \tag{1} \end{align*} $$
AT Q. 14. Answer the following :
(i) Why are the connections between the resistors in a meter bridge made of thick copper strips ?
(ii) Why is it generally preferred to obtain the balance point in the middle of the metre bridge wire?
(iii) Which material is used for the metre bridge wire and why?
Ans. (i) This is to ensure that the connections do not contribute any extra, unknown, resistances in the circuit.
(ii) This is done to minimize the percentage error in the value of the unknown resistance.
Alternatively, This is done to have a better “balancing out” of the effects of any irregularity or non-uniformity in the meter bridge wire.
This can help in increasing the sensitivity of the meter bridge circuit.
1
(iii) Manganin / Constantan / Nichrome
These materials have a low temperature (any one ) coefficient of resistance / high resistivity. $\quad 1 / 2+1 / 2$
[CBSE Marking Scheme 2014]
Long Answer Type Questions
(5 marks each)
