MAE140 HW3 Solutions

Transcription

1 MAE40 HW Solutions.7) Method: Remove load resistor and find Thevenin equivalent circuit VOC To find Isc, use a voltage divider at a a Isc Isc 00 6*00 00 Amps Rt Voc / Isc / /00 600

2 MAE40 HW Solutions.7)Continued Thevenin Equivalent Now put RL back to find power delivered Without Attenuator: V V 600/400 = 9: Power decreases by /9 Power Power V 6 V R V R 6*600 4*600 Another voltage divider Watts 600 Watts 400 In db Log Log 9. 54dB For db calculation, use 0Log(Pa/Pb), not 0Log(/) (Because power is proportional to V^, we use 0 Log(A/B) for Voltage or Amplitude comparisons, and 0Log (A/B) for Power comparisons

3 MAE40 HW Solutions.7) Initial guess:50 Ohm Voltage Divider Check boundaries, Let RL = infinity. Then I = 0 which satisfies I<= 00 ma VL = *(50/50) = 4V which also satisfies VL<= 4V. Note any decrease in RL would only lower VL so the voltage requirement will remain satisfied. Let RL = 0 I = /00 = 0 ma>00ma. We need to reduce this current So add another series 50 Ohm to do this. This will also reduce our voltage across the load, maintaining VL<=4V Check: Vmax (when RL = infinity) = *(50/00) = V Imax (When RL = 0) = /50 = 0 ma

4 MAE40 HW Solutions.76) Team A I P P Loss Error 0V 4050 I R I Team B R 5.45mA k 7mW mW *00.4% It is sometimes convenient to convert to the Thevenin equivalent * 0* 4. 7V Find : P P Loss Error R R 6 5.4W.9 5.9mW *00.4% Team A clearly wins, they have ) A closer tolerance ) They use standard resistor values (6 Ohm isn t standard) ) They waste less power (or require less power from the source) 4

5 MAE40 HW Solutions.) (Refer to circuit drawing from HW, pg ) Note this circuit is linear with just resistors and independent current sources, allowing us to directly write the equations by inspection (see T,R &T pg 75) A)Node A: (/0 + /0 + /) / = A Node B: -/ + (/4+/) = A-A Simplifying and putting into matrix form we get Computing the determinant we get 40 0 * 40 *.075 B) Vx = = -V, Ix = /0 = 0/0 = 0.5 Amps

6 .) A) 4k k k 4k Vx Vx k k MAE40 HW Solutions 0mA Vx k 0 0mA 0mA 0 Vx B) To find and, we can Node A combine the two k resistors in series to remove a node equation, then once armed with, use a voltage divider equation to get Vx. Node B C) Vx Ix k * k k k k.* k V.6mA 4k k 0 4k k k 0mA 5 4k k 0mA 5 4k k 0mA 4.55V.V

7 MAE40 HW Solutions.4) See Figure -4 in HW for reference on page Node A Node C A) By inspection, (Note is known) /k + (-)/4k = ma Vc/4k + (Vc-)/4k = -ma B) Use given values to solve for and Vc (/k + /4k) = ma + 5V/4k = V Vc(/4K) = -ma + 5V/4k Vc = 0.5V C) Solve for Vx and Ix ( = 5V) Vx = -Vc =.5V Ix = (-5V)/4k + (Vc-5V)/4k = -.65mA

8 MAE40 HW Solutions A).5) Choose ground so one side of the voltage source is connected to it (in this case, let ground be connected to, yielding Vc known) Node A (-Vc)/R + (-)/R = Is Node B (-)/R4 + (Vc-)/R = /R A R4 = R = B C Since Vc is known, move it to the RHS and put node A and node B equations into matrix form R = R = B) (/R + /R4) /R4 = Is + Vc/R /R4 (/R4 + /R + /R) = -Vc/R V.V See solution - for help inverting a x matrix Vx = -Vc =.V 0V = -7.V Ix = (Vc-)/R = (0-6.6)/k =.4mA

9 -Vx+.) We have 4 unknown nodes MAE40 HW Solutions R = R = Ix A B C R4 = R = Node A: (-)/R - (-)/R = 0 Node B: (-)/R + (-Vc)/R = Is D R5 = Node C: (-Vc)/R - (Vc-Vd)/R4 = 0 Node D: (Vc-Vd)/R4 - Vd/R5 = 0 To solve this circuit for Vx and Ix, we don t need to put this into a 4x4 matrix form. Instead, all we really need to find is, and the whole thing will fall apart. To do this, combine R and R, and combine R, R4 and R5. Then we get (at node B) (-)/(R+R) + (-0)/(R+R4+R5) = Is We can now solve for directly (-5)/500 + /500 = 00mA = 7.5V Now we can compute Ix (It is flowing through R and R) as (-)/(R+R) = Ix (7.5-5)/500 = 5 ma = Ix To solve for Vx, note the current from B to C is Is-Ix = 75 ma. Now just compute the I*R voltage drop for each resistor starting from. Vc = (Is-Ix)R = mA*50 =.75V Vd = Vc (Is-Ix)R4 =.75-(75mA)*50 = 5V Vx = Vc-Vd =.75V *You can also see that Vx must be /, since half the voltage is dropped across R+R5 and the other half is dropped across R4

10 -Vx+.)continued MAE40 HW Solutions R = R = Ix A B C R = R4 = R5 = D To Find the total power dissipated in the circuit, sum the I R drops across each resistor plus the power dissipated by the voltage source (Current entering positive terminal means the voltage supply is dissipating power) Power = Ix (R+R) + (Is-Ix) (R+R4+R5) + Ix* =.05 * * *5 =.75 Watts

11 .) MAE40 HW Solutions Super Node Note that if we can either find or Vc, we automatically know the other one (ie Vc = +0V) Also, by placing the ground as shown, we know. This 4 node circuit will only require equations as there are only unknowns A R4 R B Is R R C D The super node: (the sum of all currents entering the super node must be zero), /R4 + (-)/R + (Vc-)/R + (Vc-Vd)/R = 0 Node D: (Vc-Vd)/R = is = ma Using = = V and Vc = +0 ( /R4 + /R + /R + /R ) - Vd( /R ) = ( /R +/R) - ( /R + /R ) /R - Vd/R = Is /R

12 .)continued MAE40 HW Solutions Ix Super Node A R4 R B R R C Put into matrix form and use values given Is D -Vx Vd Vd.4 Vd = Vx = 400mV Ix = -/R4 +(-)/R = 4.06mA Power supplied by = I*V = 0V*4mA = 40.6mW

13 .4) MAE40 HW Solutions Refer to circuit drawing in T,R&T pg Node Equations ) /R + (-Vd)/R6 = is ) -/R + (Vc-)/R5 = is ) -Vc/R + (-Vc)/R6 = is 4) Vd/R4 + (Vd-)/R6 = = is To find Vx we only need Vd (place reference at Vx) So use equations and 4 ( equations, unknowns, equations and don t have the node variables we are interested in, skip them) (/R + /R6) + Vd(-/R6) = Is (-/R6) + Vd(/R4 + /R6) = Is You should be able to put this into matrix form and solve it by now = 64V Vd = 56V = Vx