Electronics - Capacitors - Discussion

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?

[A]. 10 omega.gif
[B]. 100 omega.gif
[C]. 1 k omega.gif
[D]. 1 M omega.gif

Answer: Option C

Explanation:

No answer description available for this question.

M.V.Krishna said: (Sep 3, 2011)  
V=I*R;

V=40,I=40mA;

R=40V/40mA;

R=1kohm

Rita said: (Sep 12, 2011)  
What about capacitive reactance? Total impedance is required which also include Xc.

Ravi Jumwal said: (Oct 4, 2011)  
This is the total current flowing through the circuit so where the matter of Xc come from.

Sobia said: (Apr 16, 2012)  
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

Divik said: (May 19, 2012)  
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.

Qummar said: (Aug 1, 2012)  
We get Xc = 99.47 and R = 1k and we know that total impedance is sqrt (R*R+Xc*Xc) = 1.004k.

Mukesh said: (Nov 13, 2012)  
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.

Arun said: (Oct 24, 2013)  
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.

Al Ongsuco said: (Nov 11, 2014)  
How did you get the frequency?

Shubham said: (Jan 30, 2015)  
What is total circuit impedance?

Ameya Pagare said: (Jul 18, 2019)  
The value of XC is given by the passive HPF formula for Xc.

Xc=1/2*3.14*R*C.
Xc= 99.52 ohm.

Kiran V said: (Sep 19, 2020)  
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.

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