Electronics - Capacitors - Discussion

14. 

The reactance of C2 is _____ in the given circuit.

[A]. 66 omega.gif
[B]. 80 omega.gif
[C]. 199 omega.gif
[D]. 398 omega.gif

Answer: Option D

Explanation:

No answer description available for this question.

Jai said: (Jan 8, 2012)  
Xc=1/2*pi*f*c;
xc=200ohm... How its 398 ohm?

Jagat said: (Aug 20, 2012)  
= 1 / 2*3.14*400*1*10^-6.... (C2= 1 micro farad)

= 1/0.0025136

Ankita Upadhyay said: (Apr 26, 2013)  
Total c=(2*(5+1))/(2+5+1) = 1.5.

R(total) = 1/(2*3.14*1.5*10^6) = 265.15.

Now r1/(r2||r3) = (c2||c3)/c1.

Therefore r1/(r2||r3) = 6/2 = 3.
Therefore (r2||r3) = r/(r1+r2||r3) = 265.15/4 = 66.28.

Now again r2/r3 = c3/c2.
Thus r2 = (r2||r3)* (c3+c2)/c2 = 66.28 *6 /1 = 397.68 == 398.

Kencha said: (Oct 11, 2016)  
Xc = 1/2 * pi * f * c.
Xc = 1/ 2 * 3.14 * 400 * 1 * 10^-6,
Xc = 10^6 / 2 * 3.14 * 400,
Xc = 398.089,
Xc = 398 ohm.

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