Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - Filling the Blanks (Q.No. 9)
9.
The total circuit current is _____ in the given circuit.
Discussion:
3 comments Page 1 of 1.
Vandana said:
1 decade ago
Please give me the solution.
Kit said:
8 years ago
Using the impedance formula.
Xl = 2pi(240k)(30mH),
Xc = 1/2pi(240k)(36pF),
Xt = XcXl/Xl-Xc.
Solving for Z:
Z = R*Xt/sqrt(R^2+Xt^2).
Solving for It:
It = V/Z = 2.21mA.
Xl = 2pi(240k)(30mH),
Xc = 1/2pi(240k)(36pF),
Xt = XcXl/Xl-Xc.
Solving for Z:
Z = R*Xt/sqrt(R^2+Xt^2).
Solving for It:
It = V/Z = 2.21mA.
W.Maduka said:
8 years ago
xl=2πfl
=2*3.14*240*10^3*30*10^-3
=45216Ω
xc=1/2πfc
=1 ÷ 2 * 3.14 * 240 * 10^3 * 36 * 10^-3
=18430Ω
this is Parallel circuit
Vr=Vl=Vc
IR=V/R
=50/33*10^3
=0.00151515151A
IL=50/45216
=0.00110580A
IC=V/XC.
=50/18430,
=0.002712967987A,
I=√(IR^2+(IC-IL)^2),
=√(Ans^2 + (0.002712967987 - 0.00110580)^2),
=0.00220877184A,
=2.21mA.
=2*3.14*240*10^3*30*10^-3
=45216Ω
xc=1/2πfc
=1 ÷ 2 * 3.14 * 240 * 10^3 * 36 * 10^-3
=18430Ω
this is Parallel circuit
Vr=Vl=Vc
IR=V/R
=50/33*10^3
=0.00151515151A
IL=50/45216
=0.00110580A
IC=V/XC.
=50/18430,
=0.002712967987A,
I=√(IR^2+(IC-IL)^2),
=√(Ans^2 + (0.002712967987 - 0.00110580)^2),
=0.00220877184A,
=2.21mA.
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