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प्रश्न
An ac voltage V = 280 sin(100π t) volt is connected across a series LCR circuit in which R = 400 Ω, L = `5/π` H and C = `50/π` µF. Taking `sqrt(2) = 1.4`, calculate:
- Impedance of the circuit.
- Rms value of current that flows in the circuit.
- Power factor of the circuit.
संख्यात्मक
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उत्तर
Given: V = 280 sin(100π t)
V0 = 280 V
ω = 100π
(I) Inductive Reactance (XL) = ωL
= `100 pi xx 5/pi`
= 500 Ω
Capacitive reactance `(X_C) = 1/(ωC)`
= `1/(100pi xx (50/pi xx 10^-6)`
= `10^6/5000`
= 200 Ω
Net reactance (X) = XL − XC
= 500 − 200
= 300 Ω
Impedance (Z) = `sqrt(R^2 + X^2)`
= `sqrt(400^2 + 300^2)`
= `sqrt(160000 + 90000)`
= `sqrt(250000)`
= 500 Ω
(II) `V_(rms) = (V_0)/sqrt2`
= `280/1.4`
= 200 V
`I_(rms) = (V_(rms))/Z`
= `200/500`
= 0.4 A
(III) cos ϕ = `R/Z`
= `400/500`
= 0.8
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