SME
Circuit Analysis II · Problem type
Phasor form: polar and rectangular
Use this for “Write 10∠36.87° in rectangular form, and −3 − j4 in polar form.”
Diagram
A
=
√
8
2
+
6
2
=
10.
θ = 36.87°: 10∠36.87°.
This is an example. Type your own values to start a new problem.
A
Magnitude
example
θ
Angle
example · °
a
Real part
calculated
b
Imaginary part
calculated
Formulas
a
=
A
×
cos(
θ
)
8 = 10 × cos(36.8699°)
b
=
A
×
sin(
θ
)
6 = 10 × sin(36.8699°)
A
=
√
a
2
+
b
2
10
=
√
8
2
+
6
2
θ
=
atan2(
b
,
a
)
+36.8699 = atan2(6, 8)
Assumptions
•
The imaginary unit is j (j² = −1): the phasor A∠θ is the complex number a + jb.
•
θ is measured from the positive real axis, −180° to 180°; atan2(b, a) puts it in the right quadrant.
•
v(t) = A cos(ωt + θ) has the phasor A∠θ; write a sine as a cosine first, sin x = cos(x − 90°).
Step-by-step
WE KNOW
A = 10, θ = +36.8699°
FIND
real part (a), imaginary part (b)
Step 1 · Find real part (a)
Use a = A × cos(θ)
The real part is the phasor’s shadow on the real axis.
a
=
10
×
cos(36.8699°)
a
=
10
×
0.8
a
=
8
Step 2 · Find imaginary part (b)
Use b = A × sin(θ)
The imaginary part is the phasor’s height above the real axis.
b
=
10
×
sin(36.8699°)
b
=
10
×
0.6
b
=
6
Check
Put the numbers back in. Both sides should match.
8 = 10 × cos(36.8699°) ✓
6 = 10 × sin(36.8699°) ✓
10
=
√
8
2
+
6
2
✓
36.8699 = atan2(6, 8) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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