A series RLC circuit with R = 10 Ω, C = 80 μF, and L = 10 mH is connected across a 200 Hz, 15 V ac source. The total impedance, expressed in polar form, is?
Answers
Given : A series RLC circuit with R = 10 Ω, C = 80 μF, and L = 10 mH
To Find : The total impedance, expressed in polar form,
Solution:
R = 10 Ω,
C = 80 μF = 80 x 10⁻⁶ F
L = 10 mH = 10 x10⁻³ F
XL = 2πfL
= 2π (200)10 * 10⁻³
= 4π
= 12.566
Xc = 1 /2πfC
=> Xc = 1/2π(200)80 x 10⁻⁶
=> Xc = 9.947
. The total impedance = R + jXL - jXc
= 10 + j(12.566 - 9.947)
= 10 + j 2.619
The total impedance is 10 + j 2.619
| Z | = 10.337
Ф = tan⁻¹ ( 2.619 /10) = 14.676°
Polar Form = 10.337 * e^(j 14.676°)
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The value of impedance is 10 + j2.619
Explanation:
We are given that:
- Resistance "R" = 10 ohm
- Current "C" = 80 µF
- L = 10 mH
- Frequency "F" = 200 Hz
- The voltage across the circuit "V" = 15 V
Solution:
X(L) = 2 π f L
X(L) = 2π (200)10 x 10^-3 = 4π = 12.566
Xc = 1 / 2 π f C
Xc = 1/2π(200)80 x 10⁻⁶
Xc = 9.947
Total impedence = R + jXL - jXC
Total impedence = 10 + j2.619