SATHEE: Electromagnetic Induction and Alternating Current 7 Question 16

Electromagnetic Induction and Alternating Current 7 Question 16

####16. In process 2 , total energy dissipated across the resistance $E_{D}$ is

(a) $E_{D} = \frac{1}{3} \frac{1}{2} C V_{0}^{2}$

(b) $E_{D} = 3 \frac{1}{2} C V_{0}^{2}$

(d) $E_{D} = \frac{1}{2} C V_{0}^{2}$

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Answer:

Correct Answer: 16. (a)

Solution:

For process (1)

Charge on capacitor $= \frac{C V_{0}}{3}$

Energy stored in capacitor $= \frac{1}{2} C \frac{V_{0}^{2}}{9} = \frac{C V_{0}^{2}}{18}$

Work done by battery $= \frac{C V_{0}}{3} \times \frac{V}{3} = \frac{C V_{0}^{2}}{9}$

$∴$ Heat loss $= \frac{C V_{0}^{2}}{9} - \frac{C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{18}$

For process (2)

Charge on capacitor $= \frac{2 C V_{0}}{3}$

Extra charge flow through battery $= \frac{C V_{0}}{3}$

Work done by battery $= \frac{C V_{0}}{3} \cdot \frac{2 V_{0}}{3} = \frac{2 C V_{0}^{2}}{9}$

Final energy stored in capacitor $= \frac{1}{2} C \frac{2 V_{0}}{3} \overset{4 C V_{0}^{2}}{18}$

Energy stored in process $2 = \frac{4 C V_{0}^{2}}{18} - \frac{C V_{0}^{2}}{18} = \frac{3 C V_{0}^{2}}{18}$

Heat loss in process (2) =work done by battery in process (2)

-energy stored in capacitor process (2)

$= \frac{2 C V_{0}^{2}}{9} - \frac{3 C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{18}$

For process (3) Charge on capacitor $= C V_{0}$

Extra charge flown through battery $= C V_{0} - \frac{2 C V_{0}}{3} = \frac{C V_{0}}{3}$

Work done by battery in this process $= \frac{C V_{0}}{3} (V_{0}) = \frac{C V_{0}^{2}}{3}$

Final energy stored in capacitor $= \frac{1}{2} C V_{0}^{2}$

Energy stored in this process $= \frac{1}{2} C V_{0}^{2} - \frac{4 C V_{0}^{2}}{18} = \frac{5 C V_{0}^{2}}{18}$

Heat loss in process (3) $= \frac{C V_{0}^{2}}{3} - \frac{5 C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{18}$

Now, total heat $loss (E_{D}) = \frac{C V_{0}^{2}}{18} + \frac{C V_{0}^{2}}{18} + \frac{C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{6}$

Final energy stored in capacitor $= \frac{1}{2} C V_{0}^{2}$

So, we can say that $E_{D} = \frac{1}{3} (\frac{1}{2} C V_{0}^{2})$

Electromagnetic Induction and Alternating Current 7 Question 16

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Electromagnetic Induction and Alternating Current 7 Question 16

Answer:

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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