Question

A capacitor of value 4 uF charged at 50V is connected with another capacitor of value 2uF charged at
100V. in such a way that plates of similar charges are connected together. Before joining and after
joining the total energy in multiples 10 ? J will be:​

Answer 1

Answer:

Delta E = 0.67×10^-2J

Explanation:

Based on the given data:

C1 = 4uf and V1 = 50V Then, Charge capacitor, Q1 = C1V1

= 4×10^(-6) × 50V

= 2×10^(-4).

Now,C2=2uf

Q2 = C2V2

Q2= 2×10^(-6)×100C

= 2×10^(-4).

Thus common potential, V` = (C1V1+C2V2)/C1+C2

= (2×10^(-4)+ 2×10^(-4))/6×10^-6

= 400/6

Hence, heat energy lost

= QV`/2

= 2×10^-4×400/12

= 8×10^-2/12

Delta E = 0.67×10^-2J