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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.

User Icon for bobpursley bobpursley answered
11 years ago

a, b are standard formulas

c...add the reactances and resistances as complex numbers, then convert to polar form. Current=V/Z

User Icon for Explain Bot Explain Bot answered
1 year ago

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 * ฯ€ * f * L

where:
ฯ€ โ‰ˆ 3.14159
f is the frequency in Hz
L is the inductance in Henrys

2) Capacitive Reactance (XC):
XC = 1 / (2 * ฯ€ * f * C)

where:
ฯ€ โ‰ˆ 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 = โˆš(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 ฮผF = 40 * 10^(-6) F
Voltage (V) = 120 V
Frequency (f) = 60 Hz

Using the above formulas:

a) Inductive Reactance (XL):
XL = 2 * ฯ€ * f * L
= 2 * 3.14159 * 60 * 0.2
โ‰ˆ 75.398 ohms

b) Capacitive Reactance (XC):
XC = 1 / (2 * ฯ€ * f * C)
= 1 / (2 * 3.14159 * 60 * 40 * 10^(-6))
โ‰ˆ 66.548 ohms

c) Total Reactance (X):
X = XL - XC
= 75.398 - 66.548
โ‰ˆ 8.850 ohms

d) Impedance (Z):
Z = โˆš(R^2 + X^2)
= โˆš(50^2 + 8.850^2)
โ‰ˆ 51.193 ohms

e) Current (I):
I = V / Z
= 120 / 51.193
โ‰ˆ 2.348 A (Amps)

So, the answers are:
a) Inductive Reactance (XL) โ‰ˆ 75.398 ohms
b) Capacitive Reactance (XC) โ‰ˆ 66.548 ohms
c) Total Reactance (X) โ‰ˆ 8.850 ohms
d) Impedance (Z) โ‰ˆ 51.193 ohms
e) Current (I) โ‰ˆ 2.348 A (Amps)