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) $E _D=\frac{1}{2} C V _0^{2}$
Show Answer
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 $E _D=\frac{1}{3} \Big(\frac{1}{2} C V _0^{2})$
