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 1 of 4.
G.Satish Kumar said:
7 years ago
V^2 = Vr^2+(VL-VC)^2.
V^2 = 3^2+(30-18)^2,
=9+144,
V^2= 153.
V = 12.369.
V^2 = 3^2+(30-18)^2,
=9+144,
V^2= 153.
V = 12.369.
(4)
Yasir shaikh said:
7 years ago
The voltage drop across the resistor is
E = I * R.
E = 0.003 * 1000,
E = 3 volts.
The reactive volts are;
VX = VL - VC.
VX = 30 - 18.
VX = 12 volts.
The applied volts is;
Volts = √(VR^2 + VX^2),
Volts = √(3^2 + 12^2),
Volts = √(153),
Volts = 12.3693.
E = I * R.
E = 0.003 * 1000,
E = 3 volts.
The reactive volts are;
VX = VL - VC.
VX = 30 - 18.
VX = 12 volts.
The applied volts is;
Volts = √(VR^2 + VX^2),
Volts = √(3^2 + 12^2),
Volts = √(153),
Volts = 12.3693.
(2)
KAILAS JAGTAP said:
8 years ago
V=IR=3*10^-3*1000=3V,
3^2+12^2=151,
= 12.37.
3^2+12^2=151,
= 12.37.
(1)
MUKESH said:
8 years ago
Well said, Thanks, @Govind Singh Chouhan.
(1)
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.
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.
Surya said:
1 decade ago
Can anyone please difference between impedance and resistance. Can we write v = i*z (z=impedance ).
Govind Singh Chouhan said:
1 decade ago
Given data:
R = 1000R.
Vc = 18V.
VL = 30V.
I = 3mA.
We know that,
Z=sqrt((R2)+(XL-XC)2)
And XL = VL/I ==> 30/0.003 = 10K.
And XC = VC/I ==> 18/0.003 = 6K.
So know put the value of XL and XC in in formula of Z so we can say that:
Z = sqrt(10002+(10K-6K)2)
Z = sqrt(1000000+16000000)
Z = sqrt(17000000)
Z = 4123 or
Z = 4.123K.
Now it applied voltage to the circuit is:
V=I*Z
V=0.003*4123
V=12.369 or 12.37V.
So applied voltage to this circuit is 12.37V.
Thanks.
R = 1000R.
Vc = 18V.
VL = 30V.
I = 3mA.
We know that,
Z=sqrt((R2)+(XL-XC)2)
And XL = VL/I ==> 30/0.003 = 10K.
And XC = VC/I ==> 18/0.003 = 6K.
So know put the value of XL and XC in in formula of Z so we can say that:
Z = sqrt(10002+(10K-6K)2)
Z = sqrt(1000000+16000000)
Z = sqrt(17000000)
Z = 4123 or
Z = 4.123K.
Now it applied voltage to the circuit is:
V=I*Z
V=0.003*4123
V=12.369 or 12.37V.
So applied voltage to this circuit is 12.37V.
Thanks.
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.
Vinit said:
1 decade ago
E = I*R.
E = 0.003*1000.
E = 3 volts.
The reactive volts are:
VX = VL - VC.
VX = 30 - 18.
VX = 12 volts.
The applied volts is:
Volts = Sqrt(VR^2 + VX^2).
Volts = Sqrt (3^2 + 12^2).
Volts = Sqrt(153).
Volts = 12.3693.
E = 0.003*1000.
E = 3 volts.
The reactive volts are:
VX = VL - VC.
VX = 30 - 18.
VX = 12 volts.
The applied volts is:
Volts = Sqrt(VR^2 + VX^2).
Volts = Sqrt (3^2 + 12^2).
Volts = Sqrt(153).
Volts = 12.3693.
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