current-electricity Question 59
Question: Q. 1. (i) State the working principle of a potentiometer with help of the circuit diagram, explain how the internal resistance of a cell is determined.
(ii) How are the following affected in the potentiometer circuit when (i) the internal resistance of the drive cell increases and (ii) the series resistor connected to the driver cell is reduced? Justify your answer.
A [Delhi Compt. I, II, III 2017]
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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,
$$ \begin{aligned} & I_{1} R_{1}=I_{2} R_{2} \ & \frac{V_{d 1}}{V_{d 2}}=1 \end{aligned} $$
$$ \begin{aligned} V & =I_{1} R_{1}=n e A_{1} V_{d 1} \frac{\rho l}{A_{1}} \ & =n e \rho V_{d 1} l \ V & =I_{2} R_{2}=n e \rho V_{d 2} l \end{aligned} $$
