Practice Exam #3 Prob 4 solution
1250
EX:
Find the numerical value of the
equivalent impedance, zeq, for the
circuit. Frequency ω = 1 Mr/s. Express
your answer in both rectangular and
polar form.
SOL'N:
We convert to the frequency domain by computing impedance values.
jω L = j(1M)(2 m)Ω = j2 kΩ
1
1
j
=
Ω=−
= − j2 kΩ
jω C j(1M)(500 p)
500 µ
Our frequency domain circuit:
We treat impedance just as we would resistance. We have R1 in parallel
with C and L in parallel with R2.
zeq = 1kΩ || − j2 kΩ + j2 kΩ || 1kΩ = 1kΩ ⋅ (1 || − j2 + j2 || 1)
or
⎛ 1(− j2) j2(1) ⎞
⎛ − j2
j2 ⎞
zeq = 1kΩ ⎜
+
= 1kΩ ⎜
+
⎟
⎝ 1 − j2 j2 + 1 ⎠
⎝ 1 − j2 1 + j2 ⎟⎠
We can use a common denominator and rationalize at the same time, since
the denominators are complex conjugates.
⎛ − j2 1 + j2
j2 1 − j2 ⎞
zeq = 1kΩ ⎜
⋅
+
⋅
⎝ 1 − j2 1 + j2 1 + j2 1 − j2 ⎟⎠
or
4 + j2 ⎞
⎛ 4 − j2
⎛ 4 − j2 + 4 + j2 ⎞
zeq = 1kΩ ⎜ 2
+ 2
= 1kΩ ⎜
⎟⎠
⎟
2
2
⎝
⎝1 + 2
5
1 +2 ⎠
or
⎛ 8⎞
zeq = 1kΩ ⎜ ⎟ = 1.6 kΩ
⎝ 5⎠
The answer turns out to be real because the two impedances in series are
complex conjugates of each other.
Expressing our value in rectangular (first) and polar form (second), we
have the following expressions. (Note that writing zeq = 1.6 kΩ is, in a
sense, both a rectangular and polar form and is perfectly acceptable.)
zeq = 1.6 kΩ + j0 kΩ
or
zeq = 1.6 kΩ∠0°
Note that the units are often placed only at the end of the value rather than
inside. The author prefers the above notation that keeps units closer to
numerical quantities.