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# I have a doubt in question GOHAN from competition Quick Code

 0 We had to find V(in)/V(out): V(in)=Impedance*Current The circuit given is RLC circuit with varying frequency of VOLTAGE(V(in)). The correct solutions of the question considered Impedance= Resistance +R(l)+R(c) Now that's the thing I don't understand Isn't Impedance = sqrt(r*r+(R(l)-R(c))^2) Giving V(in)/V(out)=sqrt(r*r+(R(l)-R(c))^2)/R(c) now this equation will have it's minima at s=1/sqrt(R(l)*R(c)) Resulting in itf= r/R(c)=r*sqrt(c/l) I don't understand where I am going wrong. asked 13 Jun '18, 16:56 58●3 accept rate: 33%

 0 @hrishabh15 Sorry, I can't reply to the comment since I am going to attach a screenshot. However, this was the exact doubt I had when I saw the problem. I can't understand, how such an ambiguous problem did appear in a rated contest. answered 13 Jun '18, 17:50 946●1●15 accept rate: 38% Yeah that's the thing, How can they say "They can't comment on AC or DC". They clearly mentioned some varying S, Definitely it has has to be omega. (13 Jun '18, 17:56) Yep, exactly.... I thought I am losing my grey cells over time... XD (13 Jun '18, 17:58) XDXD.. yeah I know, Poorly framed question. (13 Jun '18, 17:59) Anyway Is there a way to report it or something?? (13 Jun '18, 18:01) I don't know... I think you should first notify the @admin or the moderators before reporting straight away. (13 Jun '18, 18:05)
 0 your approach is wrong try using KVL(Kirchoff's voltage law) vin=i(R+sL+1/(s*C)) vout=i(1/(sC)) vin/vout=LCss+RC*s+1 which is a quadratic in s if a = LC and b=RC and c=1 unique value is at s= -b/(2*a) [Knowledge of quadratic equations] so s=-R/(2*L) value of vin/vout at s=-R/(2L) is 1-(RRC/(4L)) This 1-(RRC/(4*L)) is the output answered 13 Jun '18, 17:31 2★babu1998 11●1 accept rate: 0% That's an AC circuit. I don't think you can add Inductive and Capacitive Reactance with Resistance. (13 Jun '18, 17:40) You can actually. Here Voltage source is considered as RMS (it is generally the case unless specified an AC source). So you can treat inductors and capacitors as equivalent resistors and apply all DC properties here. (14 Jun '18, 01:09) Ok!! Considering Voltage is Dc source, then isn't it a open circuit because of Capacitor. (14 Jun '18, 02:10)
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question asked: 13 Jun '18, 16:56

question was seen: 191 times

last updated: 14 Jun '18, 02:10