Question

A capacitor of capacitance 'C' is initially charged to a potential difference 'V' volt. now it is connected to a battery of 2V volt with opposite polarity . the ratio of heat generated to the final energy stored in the capacitor will be?

Answer 1

^_^ ,Look for Attachment for charge Distribution in Capacitor .
This will certainly Boost your thinking, (•_^)

Final Answer : 2.25

Steps:
1) Initially, Capacitor will be charged to chage q(i) = CV on one side of plate capacitor .

After that, on changing the polarity of Battery and of EMF 2V ,Final charge on same side of plate capacitor q(f) = -C (2V) .

2) Hence, magnitude of charge transferred by Battery of 2V is
| CV - (-2CV)| = 3CV

Now,
Work done by battery = Charge transferred * EMF
= (3CV) *(2V) = 6CV^2

3) Change in Electrostatic Energy of Capacitor,
del(U) =
$\frac{1}{2} c {v(f)}^{2} - \frac{1}{2} c {v(i)}^{2} \\ \\ = \frac{1}{2} c(( {2v)}^{2} - ( {v)}^{2} ) = \frac{1}{2} c(3 {v}^{2} )$

4)
W(B) = del(U) + Heat
Don't fear , this just Energy Conservation •_^

$= > 6c {v}^{2} = \frac{3}{2} c {v}^{2} + heat \: \\ = > heat \: = \frac{9}{2} c {v}^{2}$

5) Now,
Ratio of Heat Generated to Final Energy of Capacitor = Heat / Final Energy of Capacitor

$= \frac{ \frac{9}{2}c {v}^{2} }{ \frac{1}{2} c \times {(2v)}^{2} } = \frac{9}{4} = 2.25$
Hence, Required Ratio is 2.25

Answer 2

Answer:

thx

Explanation:

I also had the same doubt