Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 5)
5.
What is the angle theta value for a 5.6
F capacitor and a 50-ohm resistor in series with a 1.1 kHz, 5 Vac source?

Discussion:
16 comments Page 2 of 2.
Ravi said:
1 decade ago
Z = R+jX, so tanθ = X/R X = 1/2PIfc place the given values in following equation.
Chandru said:
10 years ago
How to calculate this without a calculator?
(1)
Abulkalam Ajath said:
10 years ago
There is no chance to calculate without a calculator.
Sandy said:
8 years ago
Please explain in detail.
KIRAN V said:
5 years ago
Given: f = 1.1 kHz; VAC = 5V; R = 50Ω ; C = 5.6μ F.
XC = 1/ (2 * π * F * C).
XC = 1 / (2 * 3.14 * 1.1 * 103 * 5.6 * 10-6)
XC = 1 / 38.6848 * 10-3,
XC = 0.02584 * 10-3,
XC = 25.84.
The phase angle between the current and voltage is calculated from the impedance triangle above as:
For RC Circuit; tan (θ) = XC / R,
(θ) = tan-1 (XC / R),
(θ) = tan-1 (25.84/50),
(θ) = tan-1 (0.5168).
(θ) = 30.02˚.
XC = 1/ (2 * π * F * C).
XC = 1 / (2 * 3.14 * 1.1 * 103 * 5.6 * 10-6)
XC = 1 / 38.6848 * 10-3,
XC = 0.02584 * 10-3,
XC = 25.84.
The phase angle between the current and voltage is calculated from the impedance triangle above as:
For RC Circuit; tan (θ) = XC / R,
(θ) = tan-1 (XC / R),
(θ) = tan-1 (25.84/50),
(θ) = tan-1 (0.5168).
(θ) = 30.02˚.
Syed Shakeeb said:
4 years ago
@Kiran
Don't divide (XC/R) and calculate phase angle.
Calculate directly using (θ) = tan-1 (-XC / R).
You will get (θ) = -27.32˚.
Don't divide (XC/R) and calculate phase angle.
Calculate directly using (θ) = tan-1 (-XC / R).
You will get (θ) = -27.32˚.
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