Two capacitors of 36 µF and 3.8 µF are connected parallel and charged with a 51 V power supply. Calculate the total energy stored in the two capacitors.

 U=0.0517 J

Explanation:

Given that

The values of capacitance

C₁ = 36 µF

C₂ =  3.8 µF

We know that when capacitor are connected in the parallel then the equivalent value of the capacitance is given as

C=C₁ +C₂

Now by putting the values

C= 36 + 3.8 µF

C = 39.8 µF

The voltage difference

ΔV= 51 V

The total stored energy in capacitor is given as

Now by putting the values in the above equation we get

U=0.0517 J

Therefore the stored energy in the capacitors will be 0.0517 J.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

not too sure thinking all accelerations are pointed down