Electronics - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 91)
91.
What is the phase angle between the source voltage and current, when a 100 mH inductor with a reactance of 6 k
, and a 1 k resistor are in series with a source?
, and a 1 k resistor are in series with a source?Discussion:
5 comments Page 1 of 1.
KIRAN V said:
5 years ago
tan θ = (XC/R) ==> θ = tan^(-1) (XC/R) ==> (Phase angle for SERIES RC Circuit)
tan θ = (R/XC) ==> θ = tan^(-1) (R/XC) ==> (Phase angle for PARALLEL RC Circuit)
tan θ = (XL/R) ==> θ = tan^(-1) (XL/R) ==>(Phase angle for SERIES RL Circuit)
tan θ = (R/XL) ==> θ = tan^(-1) (R/XL)) ==> (Phase angle for PARALLEL RL Circuit)
Given: L = 100 m H; XL = 6 kΩ, R = 1 k Ω, θ =?
tan θ = (XL/R) = θ = tan^(-1) (XL/R) ==>(Phase angle for SERIES RL Circuit)
tan θ = (XL/R) = θ = tan^(-1) (6 kΩ /1 kΩ) ==> tan^(-1) (6) = 80.53 -> 81.00 °.
tan θ = (R/XC) ==> θ = tan^(-1) (R/XC) ==> (Phase angle for PARALLEL RC Circuit)
tan θ = (XL/R) ==> θ = tan^(-1) (XL/R) ==>(Phase angle for SERIES RL Circuit)
tan θ = (R/XL) ==> θ = tan^(-1) (R/XL)) ==> (Phase angle for PARALLEL RL Circuit)
Given: L = 100 m H; XL = 6 kΩ, R = 1 k Ω, θ =?
tan θ = (XL/R) = θ = tan^(-1) (XL/R) ==>(Phase angle for SERIES RL Circuit)
tan θ = (XL/R) = θ = tan^(-1) (6 kΩ /1 kΩ) ==> tan^(-1) (6) = 80.53 -> 81.00 °.
Ali Gohar said:
6 years ago
tan(-1) 6 = 1.40.
How is 81 coming?
How is 81 coming?
ABAR -4744 said:
6 years ago
Tan(θ) = X/R.
Here,
X is reactance of nductance, R resistance in series.
Here,
X is reactance of nductance, R resistance in series.
Sameer said:
1 decade ago
Since tan(angle) = Xl-Xc/R, here Xc is absent, therefore, tan(angle) = Xl/R.
i.e Angle = tan^-1(6)=80.5~81.0 degrees.
i.e Angle = tan^-1(6)=80.5~81.0 degrees.
Navdeep said:
1 decade ago
How is that found? May someone explain please?
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