A 50 ohm resistor is connected in series with a 0.2 H inductor, a 40 microFarad capacitor, and an AC Generator yielding 120 v (RMS) at 60 Hz. Find A) the inductive reactance b) the capacitive reactance and c) the total reactance, the impedance Z, and the current in Circuit, I.
a, b are standard formulas
c...add the reactances and resistances as complex numbers, then convert to polar form. Current=V/Z
To find the inductive reactance (XL), capacitive reactance (XC), total reactance (X), impedance (Z), and current (I), we'll use the following formulas:
1) Inductive Reactance (XL):
XL = 2 * π * f * L
where:
π ≈ 3.14159
f is the frequency in Hz
L is the inductance in Henrys
2) Capacitive Reactance (XC):
XC = 1 / (2 * π * f * C)
where:
π ≈ 3.14159
f is the frequency in Hz
C is the capacitance in Farads
3) Total Reactance (X):
X = XL - XC
4) Impedance (Z):
Z = √(R^2 + X^2)
where:
R is the resistance in ohms
X is the total reactance
5) Current (I):
I = V / Z
where:
V is the voltage in RMS (Root Mean Square)
Now, let's calculate the values:
Given:
Resistance (R) = 50 ohms
Inductance (L) = 0.2 H
Capacitance (C) = 40 μF = 40 * 10^(-6) F
Voltage (V) = 120 V
Frequency (f) = 60 Hz
Using the above formulas:
a) Inductive Reactance (XL):
XL = 2 * π * f * L
= 2 * 3.14159 * 60 * 0.2
≈ 75.398 ohms
b) Capacitive Reactance (XC):
XC = 1 / (2 * π * f * C)
= 1 / (2 * 3.14159 * 60 * 40 * 10^(-6))
≈ 66.548 ohms
c) Total Reactance (X):
X = XL - XC
= 75.398 - 66.548
≈ 8.850 ohms
d) Impedance (Z):
Z = √(R^2 + X^2)
= √(50^2 + 8.850^2)
≈ 51.193 ohms
e) Current (I):
I = V / Z
= 120 / 51.193
≈ 2.348 A (Amps)
So, the answers are:
a) Inductive Reactance (XL) ≈ 75.398 ohms
b) Capacitive Reactance (XC) ≈ 66.548 ohms
c) Total Reactance (X) ≈ 8.850 ohms
d) Impedance (Z) ≈ 51.193 ohms
e) Current (I) ≈ 2.348 A (Amps)