# test2.pdf - ECE 303 Sum2017 1 Test 2 w/Answers 1 Let a...

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ECE 303 - Sum2017 Test 2 w/Answers 1 1. Let a signal source v s ( t ) = 5 cos( 5000 t ) have source resistance R s = 75 Ω. The source is connected to a lossless line and load having Z in = 100 + j 50 Ω, as shown in the figure below: R L ~ R s Z in v t ( ) s . . x = 0 Compute P avg , the time-average power delivered to the load R L . DETAILED SOLUTION: Using phasors, replace the source/line/load in the above figure with a discrete impedance Z in . This simplifies the source/line/load down to an equivalent circuit having a phasor source voltage ˜ v s = ˜ v s ( t ) = 5 e j 5000 t in series with R s = 75 and Z in . A simplified block diagram is shown below: v (t) in ~ R S v (t) s ~ i (t) in ~ _ + ~ Z in Since the line is lossless, all power flowing into the line (i.e., flowing across Z in ) will be delivered to the load R L . Thus, we need only to find the P avg across Z in . Define ˜ v in = ˜ v in ( t ) as the voltage across Z in and ˜ i in = ˜ i in ( t ) as the current through Z in . Then the definition of P avg is given by P avg = 1 2 Re parenleftBig ˜ v in ˜ i * in parenrightBig ( a ) Using Ohm’s Law, the current ˜ i in is given by ˜ i in = ˜ v s R s + Z in ( b ) Similarly, using voltage division, the voltage ˜ v in is seen to be ˜ v in = bracketleftbigg Z in R s + Z in bracketrightbigg ˜ v s ( c ) Using equations ( b ) and ( c ), the product ˜ v in ˜ i * in can now be written as ˜ v in ˜ i * in = bracketleftbigg Z in ˜ v s R s + Z in bracketrightbigg ˜ v * s parenleftBig R s + Z in parenrightBig * = Z in vextendsingle vextendsingle vextendsingle ˜ v s vextendsingle vextendsingle vextendsingle 2 vextendsingle vextendsingle vextendsingle R s + Z in vextendsingle vextendsingle vextendsingle 2 ( d )

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ECE 303 - Sum2017 Test 2 w/Answers 2 Prob. 1 (cont.) Next, compute the quantities needed on the right side of ( d ). For the source voltage we have ˜ v s = 5 e j 5000 t | ˜ v s | 2 = 25 ( e ) Using R s and Z in from the problem statement, the sum of the impedances is R s + Z in = 75 + 100 + j 50 = 175 + j 50 ( f ) in which case the magnitude-squared term is given by vextendsingle vextendsingle vextendsingle R s + Z in vextendsingle vextendsingle vextendsingle 2 = 175 2 + 50 2 = 3 . 3125(10) 4 ( g ) Substituting ( f ) and ( g ) into equation ( d ) then gives ˜ v in ˜ i * in = parenleftBig 100 + j 50 parenrightBig ( 25 ) 3 . 3125(10) 4 = 2500 3 . 3125(10) 4 + j 1250 3 . 3125(10) 4 ( h ) Simplifying ( h ) and substituting the result into equation ( a ) then leads to the desired result P avg = 1 2 × 2500 3 . 3125(10) 4 = 0.0378 watts ( i )
ECE 303 - Sum2017 Test 2 w/Answers 3 2. A fixed charge Q 1 = 60 nanoCoul in free-space is located at ( x, y ) = (0 , 4) in the xy -plane and a fixed charge Q 2 = 20 nanoCoul is located at ( x, y ) = (8 , 8) in the xy -plane. A third mobile charge Q 3 = 10 nanoCoul is held by a restraining force at ( x, y ) = ( x p , y p ). The restraining force is removed at t = 0, but it is observed that Q 3 does not move. Compute ( x p , y p ). DETAILED SOLUTION: Draw the location of the charges in the xy -plane in the figure below: = 0 x = 8 x = 4 y = 8 y = 0 somewhere E on this line Q 1 = 60 nC Q 2 = 20 nC y x The mobile charge would move in response to the total E field created by Q 1 and Q 2 . However if the E field created by these two charges were zero then the released charge would not move. You

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