to much work

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+1)a)i) in book
+1)a)ii) in book
+1)a)iii) in book
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+1)b) in book
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+1)c)i) in book
+1)c)ii) in book
+1)c)iii) in book
+1)c)iv) in book
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+2)a)i) forward
+2)a)ii) reverse 
+2)a)iii) reverse 
+2)a)iv) forward
+2)b) in book
+2)c)i) in book
+2)c)ii) in book
+2)c)iii) in book
+2)c)iv) in book
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+3)a) in book
+3)b) in book
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+4)a)i) in book
+4)a)ii) I believe it would be close to 0V as the input of 100Hz is very low, and this is a high pass filter meaning only high frequency signals should pass through
+4)b) They can have a gain greater than 1, they can be stricter (have a sharper cut off, higher Q factor), than a passive filter
+4)c)i) in book
+4)c)ii) in book
+
+5) 
+- the diode must be between the mesurment middle point and point C, this is because there is 0 current flowing when C is connected to 9V
+- you can also tell this because current flows from A-B and B-A just fine
+- there is a resistor going from A to the mid point, and from B to the mid point
+- the resistor from A to the mid point must be 100 ohms, this is because there R = V/I, we can use the result from A-C (resistor through diode) to tell that V = 9 - 0.7 and the table to tell us that I = 0.083 which gives us 100 ohms
+- the resistor from B to the mid point must be 100 ohms, this is because there R = V/I, we can use the result from B-C (resistor through diode) to tell that V = 9 - 0.7 and the table to tell us that I = 0.0415 which gives us 200 ohms
+- this can be confirmed by the fact that using the table values fro A-B or B-A, we can say 9/0.3 to get the resistance or Ra + Rb which is 300 ohms, thus confirming our result from before 
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