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प्रश्न
Consider the LCR circuit shown in figure. Find the net current i and the phase of i. Show that i = v/Z`. Find the impedance Z for this circuit.
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उत्तर
In the circuit given above consists of a capacitor (C) and an inductor (L) connected in series and the combination is connected in parallel with a resistance R. Due to this combination, there is an oscillation of electromagnetic energy.
Potential across R = Potential of source
P.D. across R =Vm sin ωt
i2R = Vm sin ωt
`I_2 = (V_m sin ωt)/R` ......(I)
q1 is charge on the capacitor at any time t, then for series combination of C, L applying Kirchhoff's voltage law in loop ABEFA.
VC + VL = Vm sin ωt
`q_1/C + L (di_1)/(dt)` = Vm sin ωt
`q_1/C + L (d^2q_1)/(dt^2)` = Vm sin ωt ......(II)
Let q1 = qm sin(ωt + ϕ) .......(III)
`i_1 = (dq_1)/(dt)` = qm ω cos(ωt + ϕ) .......(IV)
`(d^2q_1)/(dt^2)` = qm ω2 sin(ωt + ϕ) ......(V)
Substitute the values of equations (III) and (V) in equation (II)
`(q_m sin(ωt + ϕ))/C - Lq_m ω^2 sin(ωt + ϕ)` = Vm sin ωt
`q_m sin(ωt + ϕ) [1/C - Lω^2]` = Vm sin ωt
At ϕ = 0,
`q_m sin(ωt + ϕ) [1/C - Lω^2]` = Vm sin ωt
`q_m [1/C - Lω^2] sin ωt` = Vm sin ωt
`q_m [1/C - Lω^2]` = Vm
`q_m = V_m/(ω[1/(Cω - Lω)]` ......(VI)
Applying Kirchhoff's junction rule as junction B, i = i1 + i1 using relation I, IV
i = `(V_m sin ωt)/R + q_m ω cos(ωt + ϕ)`
Now using relation VI for qm and at ϕ = 0
i = `[(V_m sin ωt)/R + (V_m ω cos ωt)/(ω[1/(ωC) - ωL])]`
i = `V_m/R sin ωt + V_m/((1/(ωC) - ωl)) cos ωt`
Let A = `V_m/r -= C cos ϕ` ......(VII)
B = `V_m/(1/(ωC) - ωL) = C cos ϕ` ......(VIII)
i = C cos ϕ sin ωt + C sin ϕ. cos ωt
= C [cos ϕ sin ωt + sin ϕ cos ωt]
i = C sin(ωt + ϕ)
Squaring and adding (VII), (VIII)
A2 + B2 = C2 cos2ϕ + C2 sin2ϕ
= C2[cos2ϕ + sin2ϕ]
A2 + B2 = C2
or C = `sqrt(A^2 + B^2)`
ϕ = `tan^-1 B/A = tan^-1 ((V_m)/(1/(ωC) - ωL))/((V_m)/R)`
∴ `tan phi = R/((1/(ωC) - ωL))`
∵ C2 + A2 = B2 = `(V_m^2)/R^2 + (V_m^2)/((1/(ωC) - ωL)^2)`
C = `[(V_m^2)/R^2 + (V_m^2)/((1/(ωC) - ωL)^2)]^(1/2)`
∵ i = `[(V_m^2)/R^2 + (V_m^2)/((1/(ωC) - ωL)^2)]_2 sin(ωt + ϕ)`
I = `V_m [1/R^2 + 1/((1/(ωC) - ωL)^2)]^(1/2) sin(ωt + ϕ)` ......(IX)
And ϕ = `tan^-1 R/((1/(ωC) - ωL))`
∵ I = `V/R` or i = `V/Z`
For ac i = `V/Z sin(ωt + ϕ)` .......(X)
Comparing (IX) and (X)
So, `1/Z = [1/R^2 + 1/((1/(ωC) - ωL)^2)]^(1/2)`
This is the impedance Z for the circuit.
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