Electronics - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 32)
32.

What is the total current for the nonideal circuit in the given circuit. If it is at resonance?

0 A
284 mu.gif A
3.3 mA
38.5 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Simpol said:   5 years ago
Leakage current in micro-ampere range.

Srideepz said:   5 years ago
Resultant current at resonance in parallel circuit I=V÷(L/CR) by sub all values we get I=286micro ampere.

Najnaj said:   7 years ago
Convert the circuit to the equivalent practical circuit.

fr = 1/2π*√(LC) = 3.2kHz.
Rpractical = ((Rs^2)+(Xl^2)) / Rs = 17612.51 ohms,
since Xl = Xc at resonance,
Z = Rpractical,

Therefore;
5/ Rpractical = 283.88 micro ampere.

Kate said:   8 years ago
Actually you solve this using Q of a practical circuit.
At resonance, Xl = Xc solving for fr,
fr = 1/2pi*sqrt(LC) = 3.2kHz.
Q = Xl/Rs = 2pi(3.2k)(75mH)/130 = 11.6.

Solving for impedance, Z:
Z = Q^2(Rs) = (11.6)^2(130) = 17491.98.

Solving for the total current, It:
It = V/Z = 5/17491.98 = 285.84 uA.
which is almost the same with 284 uA.

Hima said:   9 years ago
@Aalya.

Is z = R * (Q^2 + 1)? Please tell me, thanks in advance.

Emil Lagman said:   1 decade ago
Z = (L/RC).
It = Vs/Z.
= 286 microAmp close enough.

Ritika said:   1 decade ago
The answer should be 38.5 m by using formula v/z.

Rrgiri said:   1 decade ago
@Alya : Can't understand z = 130*(0.011^2 +1) = 130.01.

What is it's corresponding formula. Please reply, thanks in advance.

Hoha said:   1 decade ago
At resonance XL - Xc = 0.
Z = sqrt(r2) = r.
I = v/z.

A BramhaKumar said:   1 decade ago
Whats the basic formula used?


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