Recent content by oteggis

so in Polar coordinates: ##\hat n## = ##cos\theta i + sin\theta j## But I got ##\hat n## = ##-cos\theta i-sin\theta j## instead

Homework Statement Let S be the surface of a solid R , which lies inside the cylinder: ##x^2+y^2=16## and between the plane where x=0 and z=5 There is also defined a vector field F by: ##\begin{align}F(x,y)=(-x^3i-y^3j+3z^2k)\end{align}## (a) Calculate : $$\iint_{T} F.\hat n\mathrm...

This will give Area = ∫∫ 3rdrdθ using cylindrical coordinates

√x2+y2+1x2+y2+1)dA gives dS

Homework Statement Determine the area of the surface A of that portion of the paraboloid: [x][/2]+[y][/2] -2z = 0 where [x][/2]+[y][/2]≤ 8 and y≥x Homework Equations Area A = ∫∫ dS The Attempt at a Solution Area A = ∫∫ dS = 3∫∫ dS

Another Mistake, If the load is 10 ohms The load is 600 ohms and not 10 ohms

You are correct. Using the value Rtot = 10814, the gain at this stage G1 is 22 Making the total gain to be The open loop Gain G = G1*G2*G3 = (22)(157)(0.95) = 3281.3 Converting to dB: GaindB = 20 log(3281.3) ≈ 70.32dB

Sorry, it is a mistake, I wanted to write R153

Homework Statement Calculating the open loop output Impedance at 1kHz and the open loop differential gain The Attempt at a Solution Here is the approach. INPUT STAGE: Q111 and Q112 Av1 = Rtot/2re, where Rtot is the total resistance of R117 and the input resistance of Q78 which is Rin. So...