Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 9)
9.
What is the total current?
Discussion:
19 comments Page 1 of 2.
Heshan Shalanka said:
10 years ago
Guys this is how calculated this in J operation theories.
XC = 5305 and XL+18840.
(4000<0 degree)(-5305i)/((-5305i+4000<0)) = 3193.848<-37.016 degrees.
And then,
(3193.848<-37.016 degrees)(18840i)/((3193.848<-37.016 degree)+(18840i)).
This equal to 3517.1255 ohm.
And i.e. the total impedance.
So that formula of V = IR.
I = 497/3517.1255 = 0.141308 Amp = 141.308 mA.
XC = 5305 and XL+18840.
(4000<0 degree)(-5305i)/((-5305i+4000<0)) = 3193.848<-37.016 degrees.
And then,
(3193.848<-37.016 degrees)(18840i)/((3193.848<-37.016 degree)+(18840i)).
This equal to 3517.1255 ohm.
And i.e. the total impedance.
So that formula of V = IR.
I = 497/3517.1255 = 0.141308 Amp = 141.308 mA.
Vijay Bhovani said:
1 decade ago
You can't add directly the current because there is a phase differences.
We can considered the admittance.
Y= 1/R+j(WC-1/WL).
WC=1/Xc=188.4*10^-6 ohm.
1/WL=1/Xl=53.07*10^-6.
1/R=250*10^-6.
WC-1/WL=135.33*10^-6.
So total admittance,
Y=284.278*10^-6.
So,
I = V*Y = 497*284.278*10^-6 = 141.286 mA.
We can considered the admittance.
Y= 1/R+j(WC-1/WL).
WC=1/Xc=188.4*10^-6 ohm.
1/WL=1/Xl=53.07*10^-6.
1/R=250*10^-6.
WC-1/WL=135.33*10^-6.
So total admittance,
Y=284.278*10^-6.
So,
I = V*Y = 497*284.278*10^-6 = 141.286 mA.
MarkT said:
5 years ago
The answer to the choices is correct!
Xl=2piFL= 6000pi.
Xc=1/2piFC=5305.16477.
Then solve for the total impedance of RLC circuit using.
Z=1/square root (1/R^2+1/Xc^2+1/Xl^2).
Z= 3517.007 ohms.
Therefore using Ohms law,
I=V/Z= 497V/3517.007= 141.31mA.
I hope it helps!
Xl=2piFL= 6000pi.
Xc=1/2piFC=5305.16477.
Then solve for the total impedance of RLC circuit using.
Z=1/square root (1/R^2+1/Xc^2+1/Xl^2).
Z= 3517.007 ohms.
Therefore using Ohms law,
I=V/Z= 497V/3517.007= 141.31mA.
I hope it helps!
Vasu said:
1 decade ago
What lakshmi said is correct absolutely. The formula is I=Ir+Ic+Il
where in Ir=v/r
Ic=V/Xc and Il=V/Xl;
wherein Xc=1/(wc) and Xl=wl.
so we get Ir=0.12425,Ic=0.0936,Il=0.0263
So by totalling we get Itotal= 244 mA
where in Ir=v/r
Ic=V/Xc and Il=V/Xl;
wherein Xc=1/(wc) and Xl=wl.
so we get Ir=0.12425,Ic=0.0936,Il=0.0263
So by totalling we get Itotal= 244 mA
Vikash kushwaha said:
5 years ago
Let I = I1 + I2 + I3.
NOW Xl=2πfL = 2 x 3.14 x 750x4 ~ 18850ohm.
also Xc=1÷{2πfC}= 2 x 3.14 x 0.04 x 10^-6~ 5305ohm.
I= √{(497/4000)^2 + ((497/18850)-(497/5305)^2.
I= ~0.141A = 141mA.
NOW Xl=2πfL = 2 x 3.14 x 750x4 ~ 18850ohm.
also Xc=1÷{2πfC}= 2 x 3.14 x 0.04 x 10^-6~ 5305ohm.
I= √{(497/4000)^2 + ((497/18850)-(497/5305)^2.
I= ~0.141A = 141mA.
Sri said:
1 decade ago
I appreciate @Ali's answer but small correction in formula the Ix and Ic should not be individually squared and their difference should be squared inside the formula.
Elkaa said:
1 decade ago
When I calculate xl=18840 ohm, xc= 5308 xl-xc=13532 ohm
z =3836 and 497/3836 =129 ma.
please confirm where this would be wrong
best regards
thanks in advance
z =3836 and 497/3836 =129 ma.
please confirm where this would be wrong
best regards
thanks in advance
Chardice said:
10 years ago
As we analyze the circuit it is a parallel RLC so we need to get the reciprocal of R, XL, XC that is conductance, susceptance in order to get the admittance.
Deepu said:
8 years ago
You are correct @Vijay.
But Can anyone explain why can't we use formula?
Is= √((V/R)^2+((V/XL)-(V/XC))^2).
But Can anyone explain why can't we use formula?
Is= √((V/R)^2+((V/XL)-(V/XC))^2).
Lakshmi said:
1 decade ago
The answer in 244m. Because we need to add currents in each branch since its a parallel circuit! hope its noted:).
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