two inductors and a resistor
(a) The impedance of the inductor is ZL = iωL. But since ω = R/L here,
we have ZL = iR. Using the standard rules for adding impedances in
series and parallel, the total impedance of the circuit is
Z = ZL +
ZR ZL
ZR + ZL
= iR +
R(iR)
R + iR
= R
-1 + 2i
2
. (12.411)
This can also be written as Z = R(-1 + 2i)/2.
(b) The complex current is
\sqrt
I ˜ = E0/Z = E0/[R(-1 + 2i)/2] = E0(1 - 2i)/[R(-1 + 4)/2] = E0(1 - 2i)/[3R/2] = (2E0/3R)(1 - 2i) = (E0/R)√10/5 eiφ , (12.412)
where tan φ = -3. Therefore,
\sqrt
I0 =
10E0
5R
and φ = tan-1(-3) ≈ -71.6°. (12.413)
Formally,
⟨\sqrt ⟩ \sqrt
I(t) = Re I ˜ eiωt = Re
10E0
5R
eiφ eiωt =
10E0
5R
cos(ωt + φ).
(12.414)
\sqrt
(c) Since tan φ = -3 implies cos φ = 1/\sqrt10, Eq. (8.84) gives the average power dissipated in the circuit as
⟨\sqrt ⟩
1/2 E0I0 cos φ = 1/2 E0
10E0
5R
\sqrt
1
10
=
E0²
10R
. (12.415)
Alternatively, we can find the power dissipated by finding the voltage VR across the resistor and then using PR = (1/2)VR²/R. (The
resistor is the only place where power is dissipated.) The complex
voltage across the resistor equals E0 minus the complex voltage across
the upper inductor, VL. This latter voltage is
⟨ ⟩
V ˜ L = I ˜ ZL = I ˜ (iR) =
E0(1 - 2i)
R(-1 + 2i)/2
iR = E0
3 + i
5
. (12.416)
Hence
⟨ ⟩
V ˜ R = E0 - V ˜ L = E0 1 -
3 + i
5
= E0
2 - i
5
. (12.417)
\sqrt
The magnitude of this is VR = |V ˜ R| = E0/\sqrt5. Therefore,
PR =
1VR²
2R
=
1E0²/5
2R
=
E0²
10R
, (12.418)
in agreement with the above result.
734 Solutions to the problems