Question

6.
The system of capacitors connected as shown, has a total energy of 160 mj stored in it. Obtain the value of the equivalent capacitance of the system and value of Z.​

Answer 1

Answer:

Explanation:

Given The system of capacitors connected as shown, has a total energy of 160 mj stored in it. Obtain the value of the equivalent capacitance of the system and value of Z.

From the figure

5 μ f  and 5 μ f are in parallel

Now total capacitance = 10 μ f

So 5,10 and 5 μ f are in series, so equivalent capacitance

C = 1/5 + 1/10 + 1/5 = 5/10 = 1/2

  = 2 μ f

Now this capacitance is parallel with z,

C eq = 2 μ f + z μ f

Energy is given by

E = 1/2 C eq V^2

E = 160 x 10^-3 and V = 200

160 x 10^-3 = 1/2 x (2 + z) x 200^2

320 x 10^-3 = (z + 2) 10^-6 x 200^2

(z + 2) 10^-6 = 320 x 10^-3 / 4 x 10^4

z + 2 = 80 x 10^-7 / 10^-6

z + 2 = 8

z = 8 - 2

z = 6

Answer 2

Answer:

Z = 6

8 μF

Explanation:

Two capacitors 5 μF are in Parallel

Hence 5 + 5 = 10 μF

now  5μF , 10μF  , 5μF

are in series  

hence their Net  = 1/(1/5 + 1/10  + 1/5)

= 10/(2 + 1 + 2)

= 2 μF

Now 2 μF & Z μF are in parallel

Hence Equivalent capacitance = Z + 2 μF

E = (1/2) CV²

160 * 10⁻³ = (1/2) (Z + 2) * 10⁻⁶ * 200²

=> 160 * 10³ / 4 * 10⁴ = (Z + 2)/2

=> Z + 2 = 8

=> Z = 6

equivalent capacitance of the system = Z + 2 = 6 + 2 = 8 μF