SATHEE: current-electricity Question 31

current-electricity Question 31

Question: Q. 6. (i) The potential difference applied across a given resistor is altered so that the heat produced per second increases by a factor of 9 . By what factor does the applied potential difference change?

(ii) In the figure shown, an ammeter $A$ and a resistor of $4 Ω$ are connected to the terminals of the source. The emf of the source is $12 V$ having an internal resistance of $2 Ω$ . Calculate the voltmeter and ammeter readings.

R [O.D. I, II, III 2017]

Show Answer

Solution:

Ans. (i) The factor by which the potential difference changes

(ii) Ammeter Reading 1

(iii) Voltmeter Reading

$\begin{matrix} (i) & H = \frac{V^{2}}{R} \end{matrix}$

$∴ V$ increases by a factor of $\sqrt{9} = 3$ (ii) Ammeter Reading, $I = \frac{V}{R + r}$

$= \frac{12}{4 + 2} A = 2 A$

(iii) Voltmeter Reading, $V = E - I r$

$= [12 - (2 \times 2)] V = 8 V 1 / 2$

(Alternatively, $V = i R = 2 \times 4 V = 8 V$ )

[CBSE Marking Scheme 2017]

Detailed Answer :

(i) Let the original potential difference be V volts with heat generated as $H$ .

Now heat generated will be :

$\begin{matrix} (1) & H = \frac{V^{2} t}{R} \end{matrix}$

Take the new potential difference as $V^{'}$ and change in heat produced be $H$ so,

Change in heat produced :

$\begin{matrix} (2) & H^{'} = \frac{V^{' 2} t}{R} \end{matrix}$

But from the question, if the heat produced by a factor of 9 , so

$\begin{aligned} H^{'} & = 9 H \frac{V^{' 2} t}{R} & = 9 \times \frac{V^{2} t}{R} V^{' 2} & = 9 V^{2} V^{'} & = 3 V \end{aligned}$

Hence, it is clear that the applied potential difference increases by factor 3 .

[AI Q. 7. Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.

R [O.D. I, II, III 2012]

Ans. Relaxation time $(τ)$ : The average time interval between two successive collisions for the free electrons drifting within a conductor (due to the action of the applied electric field), is called relaxation time.

Relation

$\begin{aligned} v_{d} & = (- e E τ) / m & Since & i & = - n e A v_{d} & n e^{2} A τ V / m l & ∴ & V / i & = m l / (n e^{2} A τ) = ρ l / A ∴ & ρ & = m / (n e^{2} τ) \end{aligned}$

$1 / 2$

[CBSE Marking Scheme, 2012]

Current Electricity

current-electricity Question 31

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Current Electricity

current-electricity Question 31

Detailed Answer :

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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