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?
3.00 V
12.37 V
34.98 V
48.00 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
37 comments Page 4 of 4.

MUKESH said:   8 years ago
Well said, Thanks, @Govind Singh Chouhan.
(1)

KAILAS JAGTAP said:   8 years ago
V=IR=3*10^-3*1000=3V,
3^2+12^2=151,
= 12.37.
(1)

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.
(2)

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.
(4)

Bilal Niazi said:   6 years ago
Thanks @ Singh Chouhan.

Bilal Niazi said:   6 years ago
Thanks @Govind Singh Chouhan.

Abdullah said:   4 years ago
How can the KVL is not applicable in this series circuit?

Like there are 3 components in this circuit if we add 3 voltages that will equal the voltage applied according to Kirchoff's voltage law. Hence the 2 voltages of the inductor and capacitor are given(30+18=48v) and we add the resistance voltage after calculating by Ohm's law (3v) to get the total voltage of 51 volts, hence the applied voltage should also be 51 volts so as not to violate KVL.

correct me if I am wrong.


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