SME
Circuit Analysis II · Problem type
The power a load impedance draws
Use this for “A 120 V rms source feeds a load Z = 8 + j6 Ω. Find the current and the real and reactive power.”
Diagram
|Z|
=
√
8
2
+
6
2
=
10
Ω.
θ = 36.87°: Z = 10∠36.87° Ω.
The current lags the voltage by 36.87°.
This is an example. Type your own values to start a new problem.
|Z| (Ω)
=
(
R (Ω)
)
2
+
(
X (Ω)
)
2
Iᵣₘₛ (A)
=
Vᵣₘₛ (V)
|Z| (Ω)
P (W)
=
(
Iᵣₘₛ (A)
)
2
×
R (Ω)
θ
Impedance angle
calculated · °
Q
Reactive power
calculated
Formulas
|Z|
=
√
R
2
+
X
2
10
=
√
8
2
+
6
2
θ
=
tan⁻¹(
X
R
)
+36.8699
=
tan⁻¹(
6
8
)
Iᵣₘₛ = Vᵣₘₛ ÷ |Z|
12
=
120
10
P
=
Iᵣₘₛ
2
×
R
1,152
=
12
2
×
8
Q
=
Iᵣₘₛ
2
×
X
+864
=
12
2
×
6
Assumptions
•
The load is Z = R + jX (j² = −1): X > 0 is inductive, X < 0 capacitive.
•
Vᵣₘₛ and Iᵣₘₛ are rms values, the numbers an AC meter reads.
•
Only R takes power on average (P = I²R); X stores energy and hands it back each cycle (Q = I²X).
Step-by-step
WE KNOW
Vᵣₘₛ = 120 V, R = 8 Ω, X = +6 Ω
FIND
impedance (|Z|), impedance angle (θ), RMS current (Iᵣₘₛ), real power (P), reactive power (Q)
Step 1 · Find impedance (|Z|)
Use |Z| = √(R² + X²)
R
and
X
are
at
right
angles
on
the
complex
plane,
so
|Z|
is
the
hypotenuse.
|Z|
=
√
8
2
+
6
2
|Z|
=
√
64 + 36
|Z|
=
√
100
|Z| = 10 Ω
Step 2 · Find impedance angle (θ)
Use θ = tan⁻¹(X ÷ R)
The
angle
of
Z
=
R
+
jX
from
the
real
axis,
the
same
sign
as
X
.
θ
=
tan⁻¹(
6
8
)
θ
=
+36.8699°
Step 3 · Find RMS current (Iᵣₘₛ)
Use Iᵣₘₛ = Vᵣₘₛ ÷ |Z|
Ohm’s law for AC: the rms voltage over |Z|.
Iᵣₘₛ
=
120
10
Iᵣₘₛ = 12 A
Step 4 · Find real power (P)
Use P = Iᵣₘₛ² × R
The
real
power
is
heat
in
the
resistance,
the
rms
current
squared
times
R
.
P
=
12
2
×
8
P
=
144
×
8
P
=
1,152
W
Step 5 · Find reactive power (Q)
Use Q = Iᵣₘₛ² × X
The
reactive
power
is
the
rms
current
squared
times
X
,
with
the
sign
of
X
.
Q
=
12
2
×
6
Q
=
144
×
6
Q
=
+864
var
Check
Put the numbers back in. Both sides should match.
10
=
√
8
2
+
6
2
✓
+36.8699
=
tan⁻¹(
6
8
)
✓
12
=
120
10
✓
1,152
=
12
2
×
8
✓
+864
=
12
2
×
6
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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