Question

-You calculate the impedance of a single element in the circuit. The impedance is j25. What is the polar form? (25 angle -90 is incorrect, probably just the angle is incorrect not sure)

-You are using a 10x oscilloscope probe on a circuit. The alligator clip of the probe connects to the: (A is incorrect, I think it is D)

a) Vout of the circuit

b) Vin of the circuit

c) Negative side of the resistor

d) Ground reference

e) Same place as the hook clip on the probe

-Using what you learned about the Vout.Vin amplitude vs frequency log plot that you created in lab for this circuit and using what you know about the impedance of the capacitor vs frequency, what can you say about Vout for this circuit? (I think it is A or B, a high or low filter. D is incorrect)

a) It allows high frequency Vin signals to pass without changing amplitude

b) It allows low frequency Vin signals to pass without changing amplitude

c) It allows all Vin frequencies to pass equally without changing amplitude

d) It allows no Vin frequencies to pass without changing amplitude

e) It is not effected by frequency

Answer 1

Since the measured impedance is j25, therefore when we express it in polar form it becomes magnitude of 25 and having phase angle of +90. This is because real part is 0,therefore the ratio of imaginary part to real part will be +infinity because imaginary part is +ve. now inverse tangent of (+infinity) becomes +90.

Here there is no circuit probably you are finding frequency response of some simple filter.by observing the options it was like low pass filter so you have to connect the probe of oscilloscope at negative side of the resistor. Hence option c.

Since it is low pass filter and you are taking output across capacitor.it allows low frequency input signals to pass without change in amplitude. So option b.