Φ21 Fall 2006
HW9 Solutions
1 OE-41-1 Kirchho's Law Problem
For this problem, you must write and then solve the loop and node equations needed to nd the currents
I1 , I2 ,
and
I3
shown in the gure.
Solution: First the Junction equation for the top node:
∑
Iin
I2
=
∑
Iout
= I1 + I3
(1)
Figure 1: Circuit for Kirchho 's
Now the loop equation for the left loop, in a CCW direction, starting at
Law Problem
the bottom center.
−4I2 + 9 − 1I1 − 2I1 = 0
(2)
Now the loop equation for the right loop, in a CCW direction, starting at the bottom center. Notice that
this loop happens to be against the direction of the currents.
1 + 2I3 − 8 + 4I2 = 0
Use 1 to eliminate
I2
(3)
from (2):
−4 (I1 + I3 ) + 9 − 3I1
=
0
I3 =
9 7
− I1
4 4
And use this to eliminate I_2 and I_3 from (2):
(
−7 + 2
9 7
− I1
4 4
)
(
(
))
9 7
+ 4 I1 +
− I1
= 0
4 4
13
13
− I1 = −
2
2
I1 = 1 A
Now sub this back into (4) and (1) to nish.
I3 =
9 7
− I1 = 0.5 A
4 4
I2 = I1 + I3 = 1.5 A
1
(4)
2 K 31.76 Time Constant in RC Circuit
Part A. What is
∆VC
a very long time after the switch is closed? So-
lution: Once the capacitor is closed, the current is zero and only the
battery and the capacitor matter (∆VR
Part B. What is the charge
Qmax
= IR = 0),
∆VC = E .
so
that builds up on the capacitor a
Solution: The charge is
very long time after the switch has closed?
Qmax = CV = CE .
Part C. In this circuit, does
The current
I
I = +dQ/dt
or
battery. Since the capacitor starts uncharged,
0.
I = −dQ/dt?
Therefore, the current
Q is increasing, so dQ/dt >
I = dQ/dt.
Figure 2: The capacitor begins to
Part D. What is the correct expression for the current
that
τ = RC
Solution:
will ow clockwise out of the positive terminal of the
and
x
exp(x) = e
I
at time t? (Note
.) Solution: Initially, the charge on the
charge after the switch closes at
t=0
s.
capacitor is zero, so as you go around a Kirchho loop, there is only the battery and the resistor. Thus, the
I0 = E/R.
t = 0), so:
initial current is
(which is 1 at
The form of the solution is just this initial current times the decaying exponential
I(t) = I0 e−t/τ = E ·
(
)
1
t
· exp −
R
τ
3 K31.78 Energy Loss in an RC Circuit
(Note, the numbers in this problem are randomized.)
a for a very long
b for 0.2 ms, then
Ω resistor?
The switch in Figure 3 has been in position
suddenly ipped at time
t=0
to position
How much energy is dissipated by the 50
time. It is
back to
a.
Solution: The capacitor eventually charges up to the battery voltage
with position
a.
So its initial energy is
U0 =
1
1
2
CV 2 = (20 µF) (50 V) = 0.025 J
2
2
When the switch is ipped to
b,
nentially, with the time constant
the capacitor starts discharging expo-
Figure 3:
τ = RC = (50 Ω) (20 µF) = 1 ms.
K31.78.
The
current, charge, and voltage (because it's proportional to charge, i.e.
V = Q/C.)
Circuit for problem
all follow this same ex-
ponential decay. (The energy doesn't, because it's voltage squared.) After 0.2 ms, the voltage across the
capacitor is decreased, so the new energy is
U1 =
)2
(
)2
1
1
1 (
CV 2 = C V0 e−t/RC = (20 µF) 50 V e−0.2/1.0 = 0.01676 J
2
2
2
All of the missing energy is dissipated by the resistor as heat.
ER = U0 − U1 = 8.24 mJ
The energy could also be calculated by integrating the power in the resistor,
(notice the 2 in the exponential because it gets squared):
∫
ER
=
∫
tf
2
I R=R
I02 exp −2t/RC
0
2
= R (V0 /R)
=
−RC
t
[exp −2t/RC]0f
2
1
CV 2 (1 − exp −2tf /RC) = U0 − U1
2 0
2
P = IV = I 2 R
as follows
4 OE 41-4 Two capacitors, change K
(Note, the numbers in this problem are randomized.)
Two capacitors (1 and 2) are connected in parallel as shown in panel (a)
Capacitor 1 has an insulator (dielectric constant K ) that
KC10 ; capacitor
2 has only air between its plates. Each capacitor acquires a charge when
0
connected to the voltage V0 . For this problem, use the values C1 = 2 µF,
0
C2 = 5 µF, K = 4 , and V0 = 6 V.
of Figure 4.
lls the space between the plates, and its capacitance is
Parts A and B. What are the charges
Q1
and
Q2
on the capacitors for
the setup shown in panel (a)?
Solution: Capacitor 1 is modied by the dielectric, while Capacitor 2 is
not, so
Q1
Q2
(
)
= C1 V0 = KC10 V0 = 4 2 × 10−6 (6) = 48 µC
(
)
= C2 V0 = C20 V0 = 5 × 10−6 (6) = 30 µC
Parts C and D. Without discharging the capacitors, the voltage
Figure 4: Problem OE 41-4.
V0
is disconnected, and the dielectric is
removed from the rst capacitor. Then a dielectric (also dielectric constant
K = 4)
is inserted to ll the
space between the plates of the second capacitor. This nal setup is shown in panel (b). What is the charge
now on Capacitors 1 and 2?
Q01 and Q02 . Since the capacitors are separated from the
0
0
rest of the world, the total charge, Qtot = Q1 + Q2 = Q1 + Q2 = 78 µC, cannot change. The capacitors are
still in parallel, so the voltages across them must be equal:
Solution: Call the new charges on the capacitors
V1
Q01
C1
Q01
2 × 10−6
(
)
20 × 10−6 Q01
= V2
Q02
=
C2
78 × 10−6 − Q01
=
4 (5 × 10−6 )
(
) (
)
= 156 × 10−12 − 2 × 10−6 Q01
Q01
= 7.09 × 10−6 C
Q02
= 78 × 10−6 − Q01 = 70.91 µC
3