Intermediate Electronics (ITE 112-60), Spring 2018: Midterm Test : Chapters 7,8,

Question

Intermediate Electronics (ITE 112-60), Spring 2018: Midterm Test : Chapters 7,8, and 9 Na Note: Closed Book, Note book, Communication (physical or on the cloud through any media devices), o 1) What is the capacitive reactance of a 0.022 F capacitor operated at a frequency of 400 Hz? 78.7 kOhms B) 0.022 C) 18.1 Ohms D400 Ohms 2) At what rate the magnetic flux in a coil of 120 turns be changed to get a voltage of 75 Vin it? How do you use two capacitors to get a maximum increase of their total reactance? A) use only the larger capacitance B) combine in series D) combine in parallell use only the smaller capacitance long does it take to disharge a capacitor 37% of its initial voltage value on a 10 k resistor?

Explanation / Answer

1)

Capacitive Reactance = Xc = 1/2fC

Here, f = 400 and C = 0.022*10^-6

Xc =1/2*3.14*400*0.022*10^-6 = 18.09 K Ohm

So, option- 1

2)

E = N*Change in Flux

Change in Flux = E/N = 75/120 = 0.625

3)

When Capacitors are in Parallel, Ceq is less and thus reactance is more

Combine in parallel

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