Electrical Engineering - Passive Filters - Discussion
Discussion Forum : Passive Filters - General Questions (Q.No. 13)
13.
A series resonant band-stop filter consists of a 68 
resistor, a 110 mH coil, and a 0.02 
F capacitor. The internal resistance, RW, of the coil is 4 . Input voltage is 200 mV. Output voltage is taken across the coil and capacitor in series. What is the output voltage magnitude at f0?
resistor, a 110 mH coil, and a 0.02 F capacitor. The internal resistance, RW, of the coil is 4 . Input voltage is 200 mV. Output voltage is taken across the coil and capacitor in series. What is the output voltage magnitude at f0?Discussion:
3 comments Page 1 of 1.
Dhiraj j said:
8 years ago
I=200/(68+4)=2.778.
Voltage across LC = 2.778*4 = 11.08,
= 11.1mv.
Voltage across LC = 2.778*4 = 11.08,
= 11.1mv.
Mohammad Ghouse said:
1 decade ago
Since at Resonance XL = XC,
Therefore By Voltage Divider Rule,
Vo = Vi*[Rl/(Rl+R)].
= 0.2*[4/(68+4)].
= 11.1 mV.
Therefore By Voltage Divider Rule,
Vo = Vi*[Rl/(Rl+R)].
= 0.2*[4/(68+4)].
= 11.1 mV.
(2)
Texugoelectrico said:
1 decade ago
At the resonance frequency XL=XL, therefore we have:
Vo = [RL / (RL + R)] * Vi
= [4 / (68 + 4)] * 0.2
= 0.0111 V = 11.1 mV ---- B
Vo = [RL / (RL + R)] * Vi
= [4 / (68 + 4)] * 0.2
= 0.0111 V = 11.1 mV ---- B
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