Discussion :: Passive Filters - General Questions (Q.No.13)
resistor, a 110 mH coil, and a 0.02 
F capacitor. The internal resistance, RW, of the coil is 4 | Texugoelectrico said: (May 20, 2012) | |
| 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 |
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| Mohammad Ghouse said: (Dec 10, 2014) | |
| Since at Resonance XL = XC, Therefore By Voltage Divider Rule, Vo = Vi*[Rl/(Rl+R)]. = 0.2*[4/(68+4)]. = 11.1 mV. |
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| Dhiraj J said: (Sep 27, 2017) | |
| I=200/(68+4)=2.778. Voltage across LC = 2.778*4 = 11.08, = 11.1mv. |
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