University of California, Berkeley
EE 42/100
Spring 2013
Prof. K. Pister
Homework 1 Solution
1.
2 hours
(b) W = 1.8 x 3600 = 6.48 (kJ)
(c) Ampere-hours =
= 0.2 (Ah)
2.
(a)
32 Ω
8Ω
a
Req
4Ω
16 Ω
8Ω
b
32 Ω
a
Req
4Ω
16 Ω
b
32 Ω
a
Req
4Ω
8Ω
4Ω
40 Ω
b
a
Req
b
a
Req
4 Ω // 40 Ω = 3.63 Ω
b
16 Ω
(b)
3.
II1 == 33 A
A
1
I5
I4
I3
I2
4.
Start by arbitrarily labeling the all the currents in the branches and all the voltages across each resistor
12 Ω
IX
d
+ V _
5
Loop 2
3Ω
I2
c
I1
5Ω
+
V1
_
3Ω
a
+ V2 _
+ V3
+
V4
_
6Ω
_
Loop 1
15 V
10 V
b
a) Applying KVL to Loop 1:
-10 – V1 + V2 +V3 +15 = 0
Furthermore, by using Ohm’s Law, the voltages can be expressed in terms of the associated currents and
resistances. The KVL Loop1 equation can be re-written as:
-10 – (I1 * 5) + (I2 * 3) + (I2 * 3) + 15 = 0
– (5 * I1) + (6 * I2) + 5 = 0
(1)
Applying KVL to Loop 2:
-V3 – V2 + V5 + V4 = 0
-( I2*3) – (I2*3) + (-IX * 12) + (-IX * 6) = 0 ; Note the negative sign associated with IX
- 6*I2 – 18*IX = 0
Applying KCL at node C:
I1 + I2 – IX = 0
(2)
IX = I1 + I2
(3)
Substitute (3) into (2):
- 6*I2 – 18*(I1 + I2) = 0
-18*I1 – 24*I2 = 0
(4)
Simultaneously solve (4) and (1):
10
I1 = 19 = 0.5263 A
Plug I1 back into (4) yields:
I2 = - 0.3947 A
Using the values of I1 and I2 in (3):
IX = 0.1315 A
b) To find Vab, you can apply KVL to the following loop:
12 Ω
d
6Ω
3Ω
I2
c
I1
5Ω
10 V
3Ω
+ V2 _ +a + V3
+
V1
_
_
Vab
Loop
_
b
15 V
-10 - V1 + V2 + Vab = 0
Vab = 10 + V1 – V2 = 10 + (I1 * 5) - (-I2 * 3) = 13.82 V
c)
12 Ω
d
+
IX
+
V4
_
Vda
3Ω
I2
c
I1
5Ω
_
6Ω
3Ω
+ V2 _ a + V3
_
+
V1
_
15 V
10 V
b
Applying KVL, we have:
- V3 - Vda + V4 = 0
Vda = V4 - V3 = (- IX * 6) – (I2 * 3) = 0.39 V
d) Vdb = Vda + Vab = 14.21 V
e) Power associated with 10V source is P10V = I1 * V = (0.5263 A) * (10V) = 5.263 W. Power is being
absorbed by the 10V source since the value of P10V is positive. (The 15V source is actually the one
supplying the power to the circuit)
5.
The current through a resistor always goes from the positive to negative terminal of a resistor since
current goes from high to low voltage. Normally, we will choose the current direction in a branch and
that will determine the positive and negative terminals but in this case the terminals have been predetermined and so our guess for the current direction through the resistor is also set.
6.
a) Wire length L = 10 m
Wire cross-sectional area = 2mm2 = 2 x 10-6 m2
Resistivity ρ = 1.72 x 10-8 (Ω m)
𝐿
10
RCord = ρ x 𝐴 = (1.72 x 10-8) x 2𝑥10−6 = 0.086 Ω
b)
RCube
RCable
RCord
Supply
RCar
RCube
RCable
RCord
c) When there’s no current flowing, the measured voltage on the line is 124 V. This implies that the
supply voltage is also equal to 124V.
I = 12 A
RCube
+
+
Supply
RCable
RCord
VCord/Cable
_
VCube/Cord
_
RCube
RCord
RCable
VCube/Cord = 114 V, VCube/Cord = 111 V, VCar = 110 V,
RCar= VCar/I = 110V / 12A = 9.167 Ω
To find RCable, we can focus on the small portion of the circuit and apply Ohm’s Law:
+
VCar
_
RCar
I = 12 A RCable
+
+
VCord/Cable
_
VCar
_
RCar
RCable
VCord/Cable = I x (RCable + RCar + RCable) = I x (2RCable + RCar)
111 V = 12A x (2RCable + 9.167) RCable = 0.0415 Ω
To find RCord, same methodology as above can be applied:
I = 12 A
RCable
RCord
+
+
VCord/Cable
_
VCube/Cord
_
RCord
+
VCar
_
RCar
RCable
VCube/Cord = I x (2RCord + 2RCable + RCar) = I x 2RCord + I x (2RCable + RCar) = I x 2RCord + VCord/Cable
114V = 12A x 2RCord + 111V  RCord = 0.125 Ω
The measured RCord is 45% greater than what we initially calculated. The increased resistance value could
be due to the rising temperature of the wire when it conducts current. As the wire heats up (due to
Joule heating), electrons undergo more collisions, causing the resistance of the wire to also increase.
For copper, this causes a resistance increase of about 0.4% per Kelvin. In addition, the connection
between the plug and the receptacle is not perfect, and introduces some contact resistance.
To find RCube, we have:
I = 12 A
RCube
+
+
VCord/Cable
_
VCube/Cord
_
Supply
RCable
RCord
RCube
RCord
+
VCar
_
RCar
RCable
VSupply = I x (2RCube + 2RCord + 2RCable + RCar) = I x 2RCube + I x (2RCord + 2RCable + RCar) = I x 2RCube + VCube/Cord
124V = 12A x 2RCube + 114  RCube = 0.4166 Ω
c)
P= IV; V=IR
P = I2R
PCable = I2R = (12A)2 x (2 x RCable) = 11.95 W
PCord = I2R = (12A)2 x (2 x RCord) = 36 W
PCube = I2R = (12A)2 x (2 x RCube) = 120 W
7.
Standard units are:
Power = Watt [W]
Current = Ampere [A]
Voltage = Volt [V]
Energy = Joules [J]
kWh is a unit of energy.
Power = I*V = Energy/time
𝑃𝑜𝑤𝑒𝑟 =
𝐸𝑛𝑒𝑟𝑔𝑦
𝑇𝑖𝑚𝑒
𝑃𝑜𝑤𝑒𝑟 ∙ 𝑇𝑖𝑚𝑒 = 𝐸𝑛𝑒𝑟𝑔𝑦
Assuming a 24kWh battery capacity of Nissan Leaf, the required charging time with a 120V output at
12A is:
𝑇𝑖𝑚𝑒 =
𝐸𝑛𝑒𝑟𝑔𝑦
𝐸𝑛𝑒𝑟𝑔𝑦 24 ∗ 1000 𝑊 ∗ 1 ℎ𝑟
=
=
= 16.67 ℎ𝑜𝑢𝑟𝑠
𝑃𝑜𝑤𝑒𝑟
𝐼∗𝑉
12 𝐴 ∗ 120 𝑉
With a 208V outlet at 30A, the required charging time is:
𝑇𝑖𝑚𝑒 =
24 ∗ 1000 𝑊 ∗ 1 ℎ𝑟
= 3.85 ℎ𝑜𝑢𝑟𝑠
30 𝐴 ∗ 208 𝑉
8. The power delivered by a 350V 255A charger is (350V*255A)=89.25kW
The time required to charge an 85kWh battery with a 350V outlet at 255A is:
𝑇𝑖𝑚𝑒 =
85 ∗ 1000 𝑊 ∗ 1 ℎ𝑟
= 0.95 ℎ𝑜𝑢𝑟𝑠 ≈ 57 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
255 𝐴 ∗ 350 𝑉
9.
With a 20% efficiency, the maximum power that solar panel can generate is 0.02 * 1 kW/m2 = 20 W/m2.
With output voltage of 20V, the output current of one square meter of panel is:
P = I* V
I = P/V = 20 W / 20V = 1A