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) $E_{D}=3 \frac{1}{2} C V_{0}^{2}$
(c) $E_{D}=3 C V_{0}^{2}$
(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 $=\frac{C V_{0}}{3} \times \frac{V}{3}=\frac{C V_{0}^{2}}{9}$
$\therefore$ 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 $=\frac{1}{2} C \frac{2 V_{0}}{3} \quad \stackrel{4 C V_{0}^{2}}{18}$
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 $=\frac{C V_{0}}{3}\left(V_{0}\right)=\frac{C V_{0}^{2}}{3}$
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 $\operatorname{loss}\left(E_{D}\right)=\frac{C V_{0}^{2}}{18}+\frac{C V_{0}^{2}}{18}+\frac{C V_{0}^{2}}{18}=\frac{C V_{0}^{2}}{6}$
Final energy stored in capacitor $=\frac{1}{2} C V_{0}^{2}$
So, we can say that $E_{D}=\frac{1}{3} \Big(\frac{1}{2} C V_{0}^{2})$
