Question

14th one....it's urgent...............​

Answer 1

Answer:

energy stored in combination < energy stored in the single capacitor initially

Explanation:

As we know, charge (q)=cv and energy stored (u)=(1/2)cv^2

Initially, U1=(1/2)cv^2

For the combination, q1=cv1 and q2=cv2

From the conservation of charge: q = q1 + q2

cv = cv1 + cv2

v = v1 + v2

Here the combination is parallel so voltage remains constant hence, v=2V so v1=v/2 and v2=v/2

Now, U2= (1/2)cv1^2 +(1/2)cv2^2

U2= (1/2)c(v/2)^2+(1/2)c(v/2)^2

U2=(cv^2)/4

U2=[(cv^2)/2]/2

Hence U2=U1/2 So U2<U1