Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - Filling the Blanks (Q.No. 14)
14.
The reactance of C2 is _____ in the given circuit.
66 
80
199
398
Discussion:
4 comments Page 1 of 1.
Jai said:
1 decade ago
Xc=1/2*pi*f*c;
xc=200ohm... How its 398 ohm?
xc=200ohm... How its 398 ohm?
Jagat said:
1 decade ago
= 1 / 2*3.14*400*1*10^-6.... (C2= 1 micro farad)
= 1/0.0025136
= 1/0.0025136
Ankita upadhyay said:
1 decade ago
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.
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:
9 years ago
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.
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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