Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - Filling the Blanks (Q.No. 5)
5.
The output voltage at the resonant frequency is ______ in the given circuit. If a 10 k
load is connected.
Discussion:
5 comments Page 1 of 1.
KIRAN V said:
9 years ago
R1=(Rs + Rw); R2 = RL.
V0 = Vin * R2/ R1 + R2.
V0 = 6.25 * 10000/10000 + 875,
V0 = 5.97 V.
V0 = Vin * R2/ R1 + R2.
V0 = 6.25 * 10000/10000 + 875,
V0 = 5.97 V.
Paul said:
8 years ago
I think this question has the wrong answer.
Because parallel resonant circuits, has voltage as the reference, a part of that working at resonance frequency it is minimum loss and no different phase between current and voltage. I think the answer should be 6.5V.
Because parallel resonant circuits, has voltage as the reference, a part of that working at resonance frequency it is minimum loss and no different phase between current and voltage. I think the answer should be 6.5V.
Kit said:
8 years ago
@Kiran V:
Why did you use 6.25 as Vin? Its written as 6.5V.
Why did you use 6.25 as Vin? Its written as 6.5V.
Travis said:
5 years ago
First step is to obtain the resonance frequency.
1.) (fr), XL = XC --> 2*pi*fr*L = 1/ 2*pi*fr*C --> fr = 25342 Hz.
2.) XL = XC --> 923.55.
3.) (XL + R) // XC --> 11409.99<-4.64 ohms.
4.) 5333.64 // 10k --> 5333.64<-2.17.
5.) Voltage divider formula: (6.5 * 5333.64<-2.17) / (5333.64<-2.17+800).
Answer: 5.65 Volts.
1.) (fr), XL = XC --> 2*pi*fr*L = 1/ 2*pi*fr*C --> fr = 25342 Hz.
2.) XL = XC --> 923.55.
3.) (XL + R) // XC --> 11409.99<-4.64 ohms.
4.) 5333.64 // 10k --> 5333.64<-2.17.
5.) Voltage divider formula: (6.5 * 5333.64<-2.17) / (5333.64<-2.17+800).
Answer: 5.65 Volts.
Dale said:
5 years ago
Right, thanks @Travis.
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