SME
Circuit Analysis II
Phasors and AC steady state
Use this for “A series RLC circuit at 1000 rad/s has R = 30 Ω, L = 60 mH, C = 50 μF and a 100 V source. Find Z and the current.”
Diagram
X = XL − XC = 60 − 20 = 40 Ω.
|Z|
=
√
30
2
+
40
2
=
50
Ω.
θ = 53.13°: Z = 50∠53.13° Ω.
The current lags the voltage by 53.13°.
This is an example. Type your own values to start a new problem.
V
Source amplitude
example
ω
Angular frequency
example · rad/s
R
Resistance
example
L
Inductance
example
C
Capacitance
example
XL
Inductive reactance
calculated
XC
Capacitive reactance
calculated
X
Net reactance
worked out (not typed)
|Z|
Impedance magnitude
calculated
θ
Impedance angle
calculated · °
I
Current amplitude
calculated
Formulas
XL
=
ω
×
L
60 = 1,000 × 0.06
XC
=
1
ω
×
C
20
=
1
1,000 × (5 ×
10
-5
)
X
=
XL
−
XC
+40 = 60 − 20
|Z|
=
√
R
2
+
X
2
50
=
√
30
2
+
40
2
tan(
θ
)
=
X
R
tan(53.1301°)
=
40
30
I
=
V
÷
|Z|
2
=
100
50
Formulas are worked in H, F. The step-by-step shows the conversions.
Assumptions
•
The imaginary unit is j (j² = −1), as circuits write it: Z = R + j(XL − XC), the inductor’s reactance up, the capacitor’s down.
•
Steady state: every voltage and current is a sine at the source’s frequency ω; V and I are peak amplitudes.
•
The current lags the voltage by θ (leads it when θ < 0); when XL = XC the circuit is at resonance and θ = 0.
Step-by-step
WE KNOW
V = 100 V, ω = 1,000 rad/s, R = 30 Ω, L = 60 mH, C = 50 μF
FIND
inductive reactance (XL), capacitive reactance (XC), net reactance (X), impedance magnitude (|Z|), impedance angle (θ), current amplitude (I)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to H, F).
L = 60 mH = 0.06 H (1 H = 1000 mH)
C = 50 μF = 5 × 10⁻⁵ F (1 F = 10⁶ μF)
Step 2 · Find inductive reactance (XL)
Use XL = ω × L
An
inductor’s
reactance
grows
with
frequency:
ω
times
L
,
in
ohms.
XL = 1,000 × 0.06
XL = 60 Ω
Step 3 · Find capacitive reactance (XC)
Use XC = 1 ÷ (ω × C)
The capacitor’s reactance shrinks as the frequency rises: one over ωC.
XC
=
1
1,000 × (5 ×
10
-5
)
XC
=
1
0.05
XC = 20 Ω
Step 4 · Find net reactance (X)
Use X = XL − XC
jXL points up and −jXC down, so the net reactance is their difference.
X
=
60
−
20
X
=
+40
Ω
Step 5 · Find impedance magnitude (|Z|)
Use |Z| = √(R² + X²)
R
lies
along
the
real
axis
and
X
at
right
angles
to
it:
|Z|
is
the
hypotenuse.
|Z|
=
√
30
2
+
40
2
|Z|
=
√
900 + 1,600
|Z|
=
√
2
,500
|Z| = 50 Ω
Step 6 · Find impedance angle (θ)
Use tan(θ) = X ÷ R
Z
points
above
the
real
axis
(inductive):
the
current
lags
the
voltage
by
θ
.
θ
=
tan⁻¹(
X
R
)
θ
=
tan⁻¹(
40
30
)
θ
=
+53.1301°
Step 7 · Find current amplitude (I)
Use I = V ÷ |Z|
Ohm’s
law
with
the
impedance’s
size:
V
÷
|Z|.
I
=
100
50
I
=
2
A
Check
Put the numbers back in. Both sides should match.
60 = 1,000 × 0.06 ✓
20
=
1
1,000 × (5 ×
10
-5
)
✓
+40 = 60 − 20 ✓
50
=
√
30
2
+
40
2
✓
tan(53.1301°)
=
40
30
✓
2
=
100
50
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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