Solution:

Ans. (i) Statement of working principle Circuit diagram and determination of internal resistance

(ii) (a) Effect of internal resistance

(b) Series resistance

[CBSE Marking Scheme 2017]

Detailed Answer :

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

Circuit Diagram and determination : Try yourself Similar to Q. 2(i) Short Answer Type Questions-II

(ii) The question is incomplete. AII Q. 2. (i) (a) State the principle on which a potentiometer works. How can a given potentiometer be made more sensitive?

(b) In the graph shown below for two potentiometers, state with reason which of the two potentiometers, $A$ or $B$, is more sensitive.

(ii) Two metallic wires, $P_{1}$ and $P_{2}$ of the same material and same length but different cross-sectional areas, $A_{1}$ and $A_{2}$ are joined together and connected to a source of emf. Find the ratio of the drift velocities of free electrons in the two wires when they are connected (i) in series, and (ii) in parallel.

A [Foreign I, II, III 2017]

Ans. (i) (a) Principle of potentiometer How to increase sensitivity

(b) Name of potentiometer Reason

(ii) Formula

(a) Ratio of drift velocities in series

(b) Ratio of drift velocities in parallel

[CBSE Marking Scheme, 2017]

Detailed Answer :

(i) (a) Principle : Similar to Q. 2 Short Answer Type Questions-II It can be made more sensitive by decreasing current in the main circuit/decreasing potential gradient/increasing resistance put in series with the potentiometer wire.

(b) Potentiometer B has smaller value $\frac{V}{l} \quad 1 / 2+1 / 2$

(ii) In series, the current remains the same

In parallel, potential difference is same but currents are different.

Similarly,

$$ \begin{aligned} & I_{1} R_{1}=I_{2} R_{2} \ & \frac{V_{d 1}}{V_{d 2}}=1 \end{aligned} $$

$$ \begin{aligned} V & =I_{1} R_{1}=n e A_{1} V_{d 1} \frac{\rho l}{A_{1}} \ & =n e \rho V_{d 1} l \ V & =I_{2} R_{2}=n e \rho V_{d 2} l \end{aligned} $$