University of California, Berkeleyspring 2013

University of California, BerkeleySpring 2013

EE 42/100 Prof. K. Pister

Homework 1 Solution

1.

2.

(a)

(b)

3.

4.

Start by arbitrarily labeling the all the currents in the branches and all the voltages across each resistor

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(2)

Applying KCL at node C:

I1 + I2– IX = 0

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):

I1 = = 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:

-10 - V1 + V2 + Vab = 0

Vab = 10 + V1 – V2 = 10 + (I1 * 5)- (-I2 * 3) = 13.82 V

c)

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 delivered to the 10V source since the value of P10V is positive. (The 15V source is actually the one supplyingthe power to the circuit)

5.

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)

RCord = ρ x = (1.72 x 10-8) x = 0.086 Ω

b)

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.

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:

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:

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:

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

Assuming a 24kWh battery capacity of Nissan Leaf, the required charging time with a 120V output at 12A is:

With a 208V outlet at 30A, the required charging time is:

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:

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