E-84 Electric & Magnetic Circuits & Devices Tanenbaum/Wang
Solution Key for PS #1
3. Design a current divider such that I1 = 2I2, but the total resistance is 1 kW.
Answer: R1 = 1.5 kW, R2 = 3 kW
There are many ways to solve this problem. One that is particularly easy is to use the following two equations:
(1) RTotal = 1 kΩ = R1•R2/( R1 + R2) and For a current divider I1 = KR2 and I2 = KR1, so that
(2) R2/R1 = I1/I2 = 2 , i.e., R2 = 2 R1.
Substituting Eq (2) into Eq (1) we have
1 = 2R12/(3R1) = 2R1/3, so that R1 = 1.5 kΩ, and, from Eq (2), R2 = 3 kΩ.
4. a. Determine the voltage VL that would be needed at the load for a transmission line (shown below) to meet the following specifications:
RT = 1,000 W, PLoad = 109 W, PLoss = I2RT = 107 W (1% loss in transmission)
b. What is the voltage V0 that would be required at the power plant for this system?
a. We note that:
(1) PLoad = 109 = VLI and
(2) PLoss = 107 = I2RT = 103I2, so that I = 102 Amps
Substituting Eq (2) into Eq (1), we find
VL = 109/I = 107 Volts.
b. From Ohm’s Law, V0 = VL + IRT = 107 + 102•103 = 1.01•107.
Note that it’s not possible to have a 10 million volt transmission line. Therefore it’s necessary to have much lower values for RT if we wish to transmit 1 GW of power with 10% power loss.
7. Calculate the equivalent resistance for the following combinations of resistors.
a.
As in the text, we represent RN as the first two resistors (each of value R) in the chain followed by RN-1 in parallel with the second of these resistors The key to solve system is to note that RN-1 looks like RN with every resistor doubled in value. Hence for a very long chain, we can say that RN-1 ≈ 2 RN (rather than just RN as in the example in the text). Hence
RN = R + R||RN-1 ≈ R + R||2RN = R + 2RRN/((R + 2RN)
Multiplying through by R + 2RN, and solving the quadratic equation for RN, we find
b.
As in the cube with all resistors of equal value, the current is divided symmetrically (or added) at each node, so that the initial current I is divided a the each node #1 into I/3 for each leg, and then into I/6 after the next node (#2), and then added back to I/3 at each node #3, before combining to reach I at the last node.
Working backward, from V = 0, we use Ohm’s Law to have
· V = IR/3, at each node #3,
· V = IR/3 + (I/6)•2R = 2IR/3 at each node #2, and
· V = 2IR/3 + IR/3 = IR.
This shows that the equivalent resistance is R!!
c.
NOTE first that from symmetry the voltages at a and b are equal!! Hence the resistor at R between a and b is irrelevant and can be neglected. This leaves 3 paths, each with resistance 2R between the input on the top left and the output at the bottom left. For three equal resistors (2R) in parallel, the total resistance is 2R/3.
8. Given the R-2R ladder shown below:
a. Use KCL to find the voltage at each node on the top of the ladder and each of the currents.
Let A = the node above I1 and B = the node above I2. and apply KCL:
At A:
At B:
Solving, we find VA = 50 V and VB = 25 V
Using these results we easily find
I = 50/4 =12.5 A
I1 = 50/8 = 6.25 A
I2 = I3 = 25/8 = 3.125 A
b. Calculate the equivalent resistance Req directly, and verify that it leads to the same value for I.
Starting with the loop at the right, note that there are two 8 Ω resistances in parallel, so that these can be replaced by a single 4 Ω resistor (between node B and ground.
Using this result, the middle loop now has two 8 Ω resistors in parallel, so that it can be replaced by an equivalent 4 Ω resistor (between node A and ground).
Using this result, the total resistance of the ladder is just 8 Ω—the Equivalent Resistance.
Replacing the entire ladder with an equivalent 8 Ω resistor, I = 100/8 = 12.5 A in agreement with our previous result.
9. Solve for INL and VNL in the circuit shown, where VNL is related to INL by the curve on the left.
To construct the load line, we use KCL at the node (marked VNL in the figure):
(100 - VNL)/10 = (VNL/10) + INL
Solving this for INL, we find the load line equation:
INL = 10 - VNL/5
This is a straight line with values INL = 10 (for VNL = 0) and VNL = 50 (for INL = 0).
Drawing this load line on the figure, the intersection of the load line and the I vs V curve for the non linear device is at the point: VNL ≈ 20 V, INL ≈ 6 A.
(This solution also satisfies the load line equation.)
11. A major problem for the US is the need to imported oil, in some cases from countries that are unstable or unfriendly. The US imports about 10 million barrels of oil per day (MBPD) and this represents a huge transfer of wealth to the oil exporting nations. The US currently has about 250 million gasoline-powered passenger vehicles that are each driven about 12,000 miles/year and get on average 20 miles per gallon.
a. Calculate the annual cost of gasoline (at $3.00 per gallon) to drive a car 12,000 miles at 20 miles/gallon, and the amount of CO2 that is emitted (at 20 lbs per gallon).
Gal of gasoline/yr = 12,000 mi/20 mi per gal = 600 gal
Cost = $3•600 = $1800 per year.
CO2 emissions = 20•600 = 12,000 lb or 6 tons per yr.
b. Calculate the annual cost for electricity (at 10 cents/kWh) to drive an electric vehicle (EV) 12,000 miles that gets 2 miles/kWh (including losses in the battery charger), and the amount of CO2 that is emitted (at 1.2 lbs per kWh).
kWh of electricity/yr = 12,000 mi/2 mi per kWh = 6,000 kWh
Cost = $.1•6,000 = $600 per year
CO2 emissions = 1.2•6,000 = 7,200 lb or 3.6 tons per yr.
c. How long does it take to pay back the added cost for an EV that costs $10,000 more than a comparable gasoline powered vehicle?
With these figures, the EV saves $1,200 per year. Hence an added purchase price of $10,000 is repaid in savings over 10,000/1,200 = 8.3 years.
d. If one barrel (Bbl) of oil produces 42 gallons of gasoline, how many MBPD can be saved if the US replaces its entire fleet of 250 million passenger vehicles with EV’s?
Gasoline used per day = 2.5•108 vehicles•12,000 gal/yr
e. How many trillion kWh of electricity would be consumed by 250 million EV’s?
f. The US currently produces 4 trillion kWh of electricity. Discuss briefly the impact on the electrical power system of adding 250 million EV’s.
--Switching the entire fleet of passenger vehicles to EV’s would increase electricity consumption by about 38%.
--Much of the electricity for charging the EV’s would be drawn over-night, when there is a huge amount of excess capacity. However, a 38% increase in overall electricity demand would certainly require a huge investment in additional power plants and transmission lines.
--The use of distributed power from solar panels or fuel cells could reduce the need for transmission lines to some extent.
--Since the current fleet of gasoline vehicles will take many years to phase out, the demand for electricity from EV’s would increase gradually, so that the electrical power system in the country would have plenty of time to meet this demand.
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Sam Tanenbaum/Ruye Wang 1 E-84 spring, 2011