ECE 307 – Exam 1
Fall 2003
Name (print):
September 27, 2003
This test is closed book and closed notes. You may use the single equation sheet you
have hopefully brought to class. SHOW ALL WORK. Partial credit may be given.
1
Energy question (20 points)
Suppose that the voltage across two 1H inductors, in parallel, is the v(t) (shown below).
Calculate the total current, i(t), through both inductors as a piecewise function of time, for
all time. (−∞ < t < ∞).


 0,


−∞ < t < 0
 ?
2
v(t) =  t , 0 ≤ t < 2
i(t) =  ?


0, 2 ≤ t < ∞
?
HINT: Integration!
Since these inductors are in parallel, we can combine then to a single equivalent inductor.
1∗1
= 12 H. Now
Inductors add in parallel just like resistors. This implies that the, Leq = 1+1
that this reduction has been made, you can then apply the equation
iL (t) =
1 Zt
v(t0 )dt0
Leq −∞
Integrating the piecewise v(t) w.r.t. time, it naturally follows that
i(t) =





0,
2 3
t,
3
2 3
2 = 16
,
3
3
−∞ < t < 0
0≤t<2
2≤t<∞
Calculate the total energy stored in both inductors, WTLeq , at t = 10.
The total energy stored in both inductors has to be the same as the energy stored in the
equivalent inductor. Total energy stored in an inductor is given by WL (t) = 12 Li2 (t) So at
time, t = 10, i(10) = 16
. This implies that the total energy stored in both inductors is:
3
1 1
WTLeq (10) =
2 2
1
16
3
2
=
162
36
Suppose that all the energy stored in both inductors at time t = 10 is somehow transfered to a single, initially uncharged, 1F capacitor. What would be the voltage, VC , across
the capacitor after the transfer?
For a capacitor, the total energy is given by: WC (t) = 12 Cv 2 (t). Since all the energy is
2
transfered, then 16
= 12 Cv 2 (t) which, substituting in for C and solving for V , you get that
36
16
VC = √18
2
2
Charge and power question (20 points)
A 20 Volt light is rated at 1000 Watts.
2.1
Charge
What total charge flows through
the light in 1 hour?
R t1
Since i = dq
,
then
q
=
i(t)dt.
Therefore you need to know i(t). Since P = V I, this
t0
dt
implies that I = 1000/20 = 50. i(t) turns out to be a constant, therefore the integration
results in:
q = 50
Z
t1
t0
1dt = 50
Z
1hr
1dt = 50
0
C
sec
∗ 1hr ∗ 602
= 50 ∗ 602 C
sec
hr
.
What is the cost of 10 days of operation at 10 dollars per kilowatt-hour?
Cost = 10days ∗ 10
dollars
hrs
∗ 24
∗ 1KW = 2400dollars
KW H
day
3
2 ohms
4 ohms
a
b
c
i1
i2
+
32 V
i3
8 ohms
I1
+
20 V
I2
d
Figure 1: Schematic for problems 3 and 4
3
Mesh or node voltage problem (20 points)
Determine the net current through each of the resistors in the following circuit using either
method you like. First set up the problem by writing your KVL or KCL equations. Then
solve. Remember to specify your answers according to the directions given in the schematic,
i.e. i1 , i2 , i3 .
Suppose we were to choose to do mesh current analysis. The two KVL equations that
would result are as follows:
−32 + 2I1 + 8(I1 − I2 ) = 0
8(I2 − I1 ) + 4I2 − 20 = 0
or if you chose node voltage:
Vb −32
2
+
Vb
8
+
Vb +20
4
= 0
Solving the linear system for I1 and I2 , you find that
I1 ≈ 9.714 , I2 ≈ 8.143
This means that
i1 = I1 ≈ 9.714 , i2 = −I2 ≈ −8.143 , i3 = I1 − I2 ≈ 1.571
After calculating the currents, calculate the power delivered to each resistor, and delivered
by each source. Be careful of your signs.
Now that we know the branch currents, we can find the power delivered to each resistor.
Since P = I 2 R, it follows that
P2Ω = i1 2 ∗ 2 ≈ 188.7 , P4Ω = I2 2 ∗ 4 ≈ 265.2 , P8Ω = i3 2 ∗ 8 ≈ 19.8
4
We also know the current through each source, and in both cases, the current is entering the
negative terminal, which implies that both sources are delivering power to the circuit. Since
P = V I, it follows that
P32V = i1 ∗ 32 ≈ 310.86 , P20V = I2 ∗ 20 ≈ 162.86
Note the the sum of the powers dissipated by the resistors is the same as the powers delivered
by the sources.
5
4
Thevenin’s equiv (20 points)
Suppose there were a resistive load, RL attached between terminals b and c (in parallel with
the 4 ohm resistor) in problem 3. Find the Thevenin’s equivalent of this circuit at nodes b, c
as seen by the load, RL . Calculate the power, PRL , delivered to the load as a function of RL .
The Thevenin resistance seen by the load is the equivalent resistance of the circuit between
nodes b and c, provided that the load is not there, and the sources have been removed. The
removal of voltage sources amounts to shorting them out. If this is done, the Rth is found
to be
Rth = 2||8||4 ≈ 1.14Ω
All that is left to find is the V th which is the voltage open circuit (removal of the load RL ,
but note that the 4 ohm resistor stays). We know what the voltage Vbc is because we know
the current that flows through it from the previous problem. It follows therefore that
Vth = Vbc = I2 ∗ 4 ≈ 32.572
The power delivered to the load depends on the voltage across it. When the circuit has
been reduced to the Thevenin equiv, there will be the single voltage source Vth and the
single resistance Rth (both calculated above) in series with the load, RL . This mean that
the voltage divider applies.
RL
VRL = Vth ∗
RL + Rth
Since P =
V2
,
R
this implies that.
PR L =
Vth ∗
2
RL
RL +Rth
RL
6
t=0
2 ohms
a
32 V
b
+
C1 = 1F
C2 = 1F
d
Figure 2: Schematic for problems 5 and 6
5
Initial and final conditions (20 points)
Suppose that the initial voltages across the capacitors, just before time t = 0 is VC1 (0− ) =
VC2 (0− ) = 12.
Find the initial current through the 2 ohm resistor, IR (0+ ) = IC1 (0+ )+IC2 (0+ ) and initial
capacitor voltages, VC1 (0+ ) and VC2 (0+ ), just after time t = 0.
Find the steady-state current, IR (∞), through the resistor, and the voltages, VC1 (∞) and
VC2 (∞) across the capacitors.
Find the transient step response current, iR (t) through the resistor, if you can, for 5
points extra credit.
Since the voltage across a capacitor can not change instantaneously,
VC1 (0− ) = VC2 (0− ) = VC1 (0+ ) = VC2 (0+ ) = 12
The initial current through the resistor is therefore
32 − 12
= 10.
IR (0+ ) =
2
Since the definition of the steady-state is that the voltages and currents are not changing, it
follows that i = C dv
= 0. Since no current flows through the capacitors, then no current is
dt
flowing through the resistor, i.e.
IR (∞) = 0
Since no current is flowing through the resistor, there is no voltage drop across it, and the
entire 32 volts is across the capacitors, i.e.
VC1 (∞) = VC2 (∞) = 32
Since capacitors in parallel add like resistors in series, the Ceq = 2F . Based on the derivation
in class of the RC step response circuit we know that:
1
i(t) = i(0+ )e− RC t =
1
V − VC (0+) − 1 t 32 − 12 − 1 t
e RC =
e 2∗2 = 10e− 4 t
R
2
7
6
Extra Credit (10 points)
Fully derive the transient step response voltage, VR (t), from the circuit in problem 5, beginning with the application of KVL around the loop.
t
−V + Ri(t) + C1 −∞
i(t0)dt0
Rt
di
R dt
+ 1c dtd −∞
i(t0)dt0
di
R dt
+ C1 i
di
1
dt + RC i
di
1
+
idt
RC
+
ln(i(t)) − ln(i(0
))
i(t)
ln i(0
+)
=
=
=
=
=
=
=
=
=
=
0
0
0
0
0
1
− RC
R t −1
0
0+ RC dt
t
1 0 − RC
t 0+
1
− RC
t
1
− RC
t
i(t)
i(0+ )
=
e− RC t
i(t)
i(t)
=
=
i(0+)e− RC t
V −VC (0+ ) − 1 t
e RC
R
R
di
dt
Rt
di
0
0+ idt0 dt
ln(i(t0))|t0+
1
1
The voltage across the resistor is
+
1
1
v(t) = i(t)R = R V −VRC (0 ) e− RC t = (V − VC (0+))e− RC t
8