Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 21)
21.
If a current of 40 mA flows through a series circuit consisting of a 0.4 
F capacitor and a resistor in series with a 4 kHz, 40 Vac source, what is the total circuit impedance?
F capacitor and a resistor in series with a 4 kHz, 40 Vac source, what is the total circuit impedance?10 
100
1 k
1 M
Discussion:
12 comments Page 1 of 2.
KIRAN V said:
5 years ago
Given: I = 40 m A; C = 0.4μF; f = 4 kHz; V = 40 Vac; ZT =?
Z = Vac /I.
Z = 40 / 40 m A = 1 * 10^3.
Z = 1 k Ω.
Z = Vac /I.
Z = 40 / 40 m A = 1 * 10^3.
Z = 1 k Ω.
Ameya pagare said:
6 years ago
The value of XC is given by the passive HPF formula for Xc.
Xc=1/2*3.14*R*C.
Xc= 99.52 ohm.
Xc=1/2*3.14*R*C.
Xc= 99.52 ohm.
Shubham said:
1 decade ago
What is total circuit impedance?
Al Ongsuco said:
1 decade ago
How did you get the frequency?
ARUN said:
1 decade ago
Xc = 99.52 OHM.
Vc = I*Xc = 3.98 V.
Vr = SQRT (Vs^2-Vc^2) = 39.8 V.
Z= SQRT (R^2+ Xc^2) = 999.99 OHM ~ 1 KILO OHM.
Vc = I*Xc = 3.98 V.
Vr = SQRT (Vs^2-Vc^2) = 39.8 V.
Z= SQRT (R^2+ Xc^2) = 999.99 OHM ~ 1 KILO OHM.
Mukesh said:
1 decade ago
As Xc=99.47;
Now,
(Vt)^2=I^2[R^2 + Xc^2];
(40)^2=(40*10^-3)^2[R^2 + (99.47)^2];
R=995.
Approximately 1k.
Now,
(Vt)^2=I^2[R^2 + Xc^2];
(40)^2=(40*10^-3)^2[R^2 + (99.47)^2];
R=995.
Approximately 1k.
Qummar said:
1 decade ago
We get Xc = 99.47 and R = 1k and we know that total impedance is sqrt (R*R+Xc*Xc) = 1.004k.
Divik said:
1 decade ago
The solution of question also states that with such alternating high frequency the capacitive reactance becomes negligible as it also the fundamental property of capacitor.
Sobia said:
1 decade ago
Reactance if we calculate Xc=1/(2*pi*f*c) is 99.47 which is very low as compared to R which is 1k by above formula explained by Krishna so we ignore 99.91 and call total resistance 1k
Ravi Jumwal said:
1 decade ago
This is the total current flowing through the circuit so where the matter of Xc come from.
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