Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 1)
1.
What is the applied voltage for a series RLC circuit when IT = 3 mA, VL = 30 V, VC = 18 V, and R = 1000 ohms?
Discussion:
37 comments Page 2 of 4.
Sukumar said:
1 decade ago
According to Lalit Garg
Givendata:-
Vl=30;Vc=18;It=0.03A;
Xl=Vl/It--->30/0.03=1000
Xc=Vc/It--->18/0.03=600
Z=(R^2+(Xc-Xl)^2)^1/2
then z=1077.03
Vsource=Z*It=32.31V
but this is not valid with options.
Givendata:-
Vl=30;Vc=18;It=0.03A;
Xl=Vl/It--->30/0.03=1000
Xc=Vc/It--->18/0.03=600
Z=(R^2+(Xc-Xl)^2)^1/2
then z=1077.03
Vsource=Z*It=32.31V
but this is not valid with options.
Weerasinghege maduka said:
9 years ago
Create voltare trangle.
v^2 = (VL-VC)^2 + VR^2
So,
VR= I * R
VR= 3mA*1000Ohms
VR=3V
V^2=(XL-XC)^2+VR^2,
=(30-18)^2+3^2,
=12^2 + 9,
=144 + 9,
=> V^2 = 153.
V = 12.36.
v^2 = (VL-VC)^2 + VR^2
So,
VR= I * R
VR= 3mA*1000Ohms
VR=3V
V^2=(XL-XC)^2+VR^2,
=(30-18)^2+3^2,
=12^2 + 9,
=144 + 9,
=> V^2 = 153.
V = 12.36.
Swathi said:
1 decade ago
XL=VL/I=10 and XC=VC/I=6
Z=sqrt of (100)2+(6-10)2=10016 i got
V=I*Z
=3*10016
=300.239904077 i got but the given answer is not come how to solve to get that answer.
Z=sqrt of (100)2+(6-10)2=10016 i got
V=I*Z
=3*10016
=300.239904077 i got but the given answer is not come how to solve to get that answer.
Swapnil said:
1 decade ago
Vr = 1000*10^-3 = 3v.
VL = 30v.
vc = 18v.
V = square root of(3^2+(30-18)^2).
= square root of (9+144).
= square root of (153).
= 12.3691.
Answer = B.
VL = 30v.
vc = 18v.
V = square root of(3^2+(30-18)^2).
= square root of (9+144).
= square root of (153).
= 12.3691.
Answer = B.
Biru said:
1 decade ago
Given voltage across inductor = 30v and capacitor=18v.
R= 1000 ohms, so voltage across "R" is 1000*3mA = 3volt.
So applied voltage = 30V + 18V + 3V = 51V.
R= 1000 ohms, so voltage across "R" is 1000*3mA = 3volt.
So applied voltage = 30V + 18V + 3V = 51V.
Hussain said:
1 decade ago
Vinodhine the value of R = 1000 that is 1K but u have taken as 100.
Galit garg is right.
I = v/z => v = I * z = 3m * sqrt(17)*k = 3 * 4.123 = 12.37v
Galit garg is right.
I = v/z => v = I * z = 3m * sqrt(17)*k = 3 * 4.123 = 12.37v
RAJKAMAL GUPTA said:
1 decade ago
From phasor diagram it is clear that VL > Vc
V = sqrt[Vr^2+(VL-Vc)^2].
On putting the given value on formula as given unit we get exact answer.
V = sqrt[Vr^2+(VL-Vc)^2].
On putting the given value on formula as given unit we get exact answer.
Vinodhine said:
1 decade ago
Swathi, you have to consider the R value as 1 Kohm, and dont forget to note the current is in milliampere so resistance will be in Kiloohms
ANSHUMAN said:
1 decade ago
Vl=30;Vc=18;It=0.03A;
Xl=Vl/It--->30/0.03=1000
Xc=Vc/It--->18/0.03=600
Z=(R^2+(Xc-Xl)^2)^1/2
then z=1077.03
Vsource=Z*It=32.31V
Xl=Vl/It--->30/0.03=1000
Xc=Vc/It--->18/0.03=600
Z=(R^2+(Xc-Xl)^2)^1/2
then z=1077.03
Vsource=Z*It=32.31V
Badar said:
1 decade ago
Because votlage across inductor and capacitor are opposite in phase about 180 degree. They cancel the effect of each other :-).
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