SME
Circuit Analysis II · Problem type
Parallel RC: impedance from admittance
Use this for “A 50 Ω resistor is in parallel with a capacitor of reactance 25 Ω. Find the impedance and its angle.”
Diagram
|Z|
=
√(
10
2
+
(−20)
2
)
=
22.36
Ω.
θ = −63.43°: Z = 22.36∠−63.43° Ω.
The current leads the voltage by 63.43°.
This is an example. Type your own values to start a new problem.
R
Resistance
example
XC
Capacitive reactance
example
G
Conductance
calculated
B
Susceptance
calculated
|Y|
Admittance magnitude
calculated
|Z|
Impedance magnitude
calculated
θ
Impedance angle
calculated · °
Rₛ
Series-equivalent resistance
worked out (not typed)
Xₛ
Series-equivalent reactance
worked out (not typed)
Formulas
Rₛ
=
|Z|
×
cos(
θ
)
10 = 22.3607 × cos(−63.4349°)
Xₛ
=
|Z|
×
sin(
θ
)
−20 = 22.3607 × sin(−63.4349°)
G
=
1
R
0.02
=
1
50
B
=
1
XC
0.04
=
1
25
|Y|
=
√
G
2
+
B
2
0.04472
=
√
0.02
2
+
0.04
2
|Z| = 1 ÷ |Y|
22.3607
=
1
0.04472
θ
=
−tan⁻¹(
B
G
)
−63.4349
=
−tan⁻¹(
0.04
0.02
)
Formulas are worked in S. The step-by-step shows the conversions.
Assumptions
•
The imaginary unit is j (j² = −1); in parallel the admittances add: Y = G + jB with G = 1 ÷ R and B = 1 ÷ XC.
•
Z = 1 ÷ Y: its size is 1 ÷ |Y| and its angle is minus Y’s, so a parallel RC has θ < 0.
•
Conductance and susceptance are in siemens (1 S = 1 ÷ Ω), shown in mS.
Step-by-step
WE KNOW
R = 50 Ω, XC = 25 Ω
FIND
conductance (G), susceptance (B), admittance magnitude (|Y|), impedance magnitude (|Z|), impedance angle (θ), series-equivalent resistance (Rₛ), series-equivalent reactance (Xₛ)
Step 1 · Find conductance (G)
Use G = 1 ÷ R
Conductance is one over resistance.
G
=
1
50
G
=
0.02
S
Step 2 · Find susceptance (B)
Use B = 1 ÷ XC
Susceptance is one over reactance.
B
=
1
25
B
=
0.04
S
Step 3 · Find admittance magnitude (|Y|)
Use |Y| = √(G² + B²)
G
is
real
and
jB
imaginary,
at
right
angles:
|Y|
is
the
hypotenuse.
|Y|
=
√
0.02
2
+
0.04
2
|Y|
=
√
0.0004 + 0.0016
|Y|
=
√
0.002
|Y| = 0.04472 S
Step 4 · Find impedance magnitude (|Z|)
Use |Z| = 1 ÷ |Y|
Impedance is one over admittance, so the sizes are too.
|Z|
=
1
0.0447214
|Z| = 22.3607 Ω
Step 5 · Find impedance angle (θ)
Use θ = −tan⁻¹(B ÷ G)
Y
points
up
at
tan⁻¹(
B
G
);
dividing
1
by
Y
flips
the
angle,
so
Z
points
down:
the
current
leads.
θ
=
−tan⁻¹(
0.04
0.02
)
θ
=
−63.4349°
Step 6 · Find series-equivalent resistance (Rₛ)
Use Rₛ = |Z| × cos(θ)
Z’s shadow on the real axis: the resistance of the series pair that would act the same.
Rₛ = 22.3607 × cos(−63.4349°)
Rₛ = 22.3607 × 0.4472
Rₛ = 10 Ω
Step 7 · Find series-equivalent reactance (Xₛ)
Use Xₛ = |Z| × sin(θ)
Z’s height below the real axis: the series-equivalent reactance, negative since it is capacitive.
Xₛ = 22.3607 × sin(−63.4349°)
Xₛ = 22.3607 × (−0.8944)
Xₛ = −20 Ω
Step 8 · Convert the answers
Change each answer back to the unit you chose.
G = 0.02 S = 20 mS (1 S = 1000 mS)
B = 0.04 S = 40 mS (1 S = 1000 mS)
|Y| = 0.04472 S = 44.72136 mS (1 S = 1000 mS)
Check
Put the numbers back in. Both sides should match.
10 = 22.3607 × cos(−63.4349°) ✓
−20 = 22.3607 × sin(−63.4349°) ✓
0.02
=
1
50
✓
0.04
=
1
25
✓
0.0447214
=
√
0.02
2
+
0.04
2
✓
22.3607
=
1
0.0447214
✓
−63.4349
=
−tan⁻¹(
0.04
0.02
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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