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EE 40 Homework 1 Solutions
July 6, 2008
1. Problem 1.7
1A = 1
C
electrons
electron
·1
= 6.25 × 1018
−19
s
1.6 × 10 C
sec
2. Problem 1.15
electron
m3
15 C
·
·
·
2
−19
10
C} | {z
s}
{z
} |1.6 × {z
|10 9 electrons
electron density of Cu charge per electron current
3. Problem 1.25
a.
1
2.05∗10−3 m 2
|
2
= (5 A)(10sin(200πt) V )
1
π
{z
cross sectional area
p(t) = i(t)v(t)
= 50sin(200πt) W
= 0.284
mm
s
}
b.
E=
Z
t1
t0
p(t) dt =
Z
t1
50sin(200πt) dt =
t0
50
[cos(200πt1 ) − cos(200πt0 )]
200π
If t1 = 5ms and t0 = 0s then E = 0.1592 J
c. Using the integral computed in part b with t1 = 10ms and t0 = 0s, we find that E = 0 J. A
quicker way to find this same answer is to see that 10ms is the period of the power function,
over which time no net power will be delivered.
4. Prolbem 1.27
Using the passive sign convention, P = IV. That is, when the signs of the voltage and the
current match, power is being dissipated by the circuit element.
a. 60W delivered
b. 60W emitted
c. 60W delivered
5. Problem 1.39
given: ia = −1A, ic = 3A, ig = 5A, ih = 1A
using KCL on node 1: ib = ia + ic = 2A
using KCL on node 5: if = ig + ih = 6A
using KCL on node 4: ie = ic + ih = 4A
using KCL on node 3: id = ig + ie − ib = 7A
6. Problem 1.43
given: va = 5V, vb = 7V, vf = −10V, vh = 6V
using KVL on loop 1: vd = va + vb = 12V
using KVL on loop 2: vg = vf + vd = 2V
2
using KVL on loop 4: ve = vh + vg = 8V
using KVL on loop 3: vc = vb − ve = −1V
7. Problem 1.61
Because the current i must flow out of the ideal current source in the direction indicated, the
direction of the current matches the direction of the voltage drop across the 10V source. Since
these are the same sign, the passive sign convention tells us that the 10V source absorbs power
and the 2A source delivers power by conservation of energy.
8. Problem 1.63
The resistor is always absorbing power because it is a passive element.
Using superposition:
Adding the two contributions to iR , we get that iR = 2A. By KCL, i0 + iR = 2A ⇒ i0 = 0A.
Therefore the voltage source is inactive and neither delivers nor absorbs power. Because energy
in the circuit must be conserved, the current source must be emitting power.
9. Problem 1.71
Using KVL on the outer loop, we get that 18V = vx + 2vx ⇒ vx = 6V
3
Use this voltage to compute i0 and i1 : i0 =
Using KCL, iy = i0 − i1 = 2A
vx
2Ω
= 3A, i1 =
2vx
12Ω
= 1A
10. Problem 2.2
Let R1 //R2 denote combining resistors R1 and R2 in parallel. Then a.
Rab = 4Ω + 2Ω + (48Ω// (6Ω + 18Ω//9Ω + 12Ω//6Ω))
= 4Ω + 2Ω + (48Ω//(6Ω + 6Ω + 4Ω))
= 4Ω + 2Ω + (48Ω//16Ω)
= 4Ω + 2Ω + 12Ω = 18Ω
b.
Rab = 4Ω + [(16Ω + 8Ω)//8Ω] + 3Ω + [(6Ω + 15Ω)//28Ω]
= 4Ω + 6Ω + 3Ω + 12Ω
= 25Ω
11. Problem 2.23
5Ω
· v1 ⇒ v1 = 30V .
The 5Ω and 25Ω resistors act as a voltage divider over v1 , so v2 = 5Ω+25Ω
The 12Ω, 6Ω and the equivalent resistance of the rest of the resistors (over v1 ) form a voltage
divider over the source voltage vs .
Req = 10Ω//(30Ω//(25Ω + 5Ω)) = 10Ω//(30Ω//30Ω) = 6Ω.
Req
6Ω
· vs = 12Ω+6Ω+6Ω
· vs
v1 = 12Ω+6Ω+R
eq
⇒ vs =
24Ω
6Ω
· v1 = 120V
12. Problem 2.40
4
a.
12V
= 100mA
R1 + R2
R2
= 5V ⇒
(12V ) = 5V
R1 + R2
5
= (R1 + R2 )
12
5
= R1
7
12V
= 120Ω
=
100mA
i0 = 100mA ⇒
v0
R2
⇒ R2
R1 + R2
12
R1 = 120Ω
7
⇒ R1 = 70Ω, R2 = 50Ω
b. Adding the 200Ω resistor across the open ports means the voltage v0 can be computed
by using the voltage divider pinciple by computing the voltage across the 50Ω and the 200Ω
resistor combined in parallel:
v0 =
40
50//200
(12V ) =
(12V ) = 4.36V
(50//200) + 70
40 + 70
13. Problem 2.52
Pick ground to be as indicated in the diagram above. From the 20V source we know that the
value of v0 is 20V. Using nodal analysis to solve for v1 , v2 , we get a system of linear equations.
At node 2:
v2 − 20 v2 − v1
+
= 2 ⇒ 2v2 − 40 + 5v2 − 5v1 − 20 = 0 ⇒ −5v1 + 7v2 = 0
5
2
At node 1:
v1 − v2 v1 − 20
+
− 3 = 0 ⇒ 5v1 − 5v2 + v1 − 20 − 30 = 0 ⇒ −6v1 + 5v2 = 0
2
10
Solving this system of equations, we find that v1 = 13.53V , v2 = 18.24V . Consequently,
1
i1 = 20−v
10 = 0.647A
14. Problem 2.60
5
Using KVL on loop 1 we know that 5ix = 10ix + v2 ⇒ ix = −0.2v2
Write the nodal equation for v2 :
v2
v2
1
= 1 + ix ⇒ 0.2v2 = 1 ⇒ v2 =
= 4V
20
20
0.2 + 0.05
Therefore, ix = −0.2(4) A = −0.8A.
Write the nodal equation for v1 :
v1 = −15 + 5ix V
= −15 + 5(−0.8) V
= −19 V
6