SATHEE: current-electricity Question 60

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Question: Q. 3. Two students $X$ and $Y$ perform an experiment on potentiometer separately using the circuit diagram shown below.

Keeping other things unchanged (a) $X$ increases the value of resistance $R$ , (b) $Y$ decreases the value of resistance $S$ in the set up. How will these changes affect the position of null point in each case and why?

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Solution:

Ans. (i) If $R$ increases, current in circuit of potentiometer wire decreases making terminal potential difference between the poles to balance against large potentiometer wire length which shifts balance point towards $B$ .

$2^{1 / 2}$

(ii) If $S$ decreases, current will increase due to Ohms law, decreasing the terminal potential difference among its poles. In such case, the balance point shifts towards point $A$ . AI Q. 4. (i) State the principle of working of a potentiometer

(ii) In the following potentiometer circuit $A B$ is a uniform wire of length $1 m$ and resistance $10 Ω$ . Calculate the potential gradient along the wire and balance length $AO (= l)$ .

A [Delhi I, II, III 2016]

Ans. (i) Principle : Try yourself Similar to Q. 2 Short Answer Type Questions-II

(ii) Current flowing in the potentiometer wire

$\frac{E}{R_{total}} = \frac{2.0}{15 + 10} = \frac{2}{25} A 1 / 2$

$∴$ Potential difference across the two ends of the wire

$V_{A B} = \frac{2}{25} \times 10 V = \frac{20}{25} = 0.8 volt 1 / 2$

Hence potential gradient, $K = \frac{V_{A B}}{l_{A B}}$

$= \frac{0.8}{1.0} = 0.8 \frac{V}{m}$

Current flowing in the circuit containing experimental cell,

$= \frac{1.5}{1.2 + 0.3} = 1 A$

Hence, potential difference across length $A O$ of the wire

$\begin{aligned} = 0.3 \times 1 V = 0.3 V \Rightarrow 0.3 & = K \times l_{AO} & = 0.8 \times l_{AO} \Rightarrow l_{A O} & = \frac{0.3}{0.8} m = 0.375 m & = 37.5 cm \end{aligned}$

[CBSE Marking Scheme 2016]

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एनसीईआरटी पुस्तकें और समाधान

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