Practice Problems: A Review of Basic Circuit Analysis Solutions

1. (easy) Explain, using the concepts discussed in the previous lecture, how the drift velocity of charges in a circuit is small in comparison to the speed of the signal that causes them to move.
The drift velocity taken up by the charges is the result of electric force exerted on them by an E-field that is set up in the conductor due to a difference in electric potential across the conductor. The E-field (the signal) is set up across the conductor almost instantly, but the force exerted on the charges causes an acceleration that results in a specific (and relatively slow) speeds when the resistance to the motion is taken into account. This resistance is determined by the internal structure of the conductor. The drift velocity can be compared to the marching of soldiers (charges) under the direction of a captain's orders to march (the signal). In this analogy, the orders to march move at the speed of sound, whereas the soldiers march at a much slower speed.

2. (moderate) If 5x10¹⁶ electrons pass a point in a conductor every 2 ms, find the current in the conductor. (Recall that each electron has a charge of 1.6x10^-19C)
I = ΔQ/Δt for a steady current
I = 5x10¹⁶(1.6x10^-19)/0.002 = 4 A

3. (moderate) How much time must pass for an electron to travel 0.85 m in a copper conductor whose cross sectional area is 0.21 cm² if the current in the wire is 300 A? Use the n value for copper as n = 8.47x10²⁸ electrons/m³.  
I = qnAv_d 
300 = (1.6x10^-19)(8.47x10²⁸)(0.21x10^-4m²)v_d 
v_d = 1.05x10^-3 m/s
v_d = Δx/Δt 
1.05x10^-3 = 0.85/Δt
Δt = 800 seconds

4. (easy) A piece of iron is shaped as a block with a length of 0.15 m, a width of 0.012 m and a depth of 0.012 m. If the the resistivity of iron is 9.7x10^-8 Ωm, find the resistance of this object along its length? Does the resistance change if measured along its depth. If so, why?
R = ρL/A
Along its length: R = (9.7x10^-8)(0.15)/[(0.012)(0.012)] = 1.0x10^-4 Ω
Along its depth: R = (9.7x10^-8)(0.012/[(0.15)(0.012)] = 6.5x10^-7 Ω 
The resistance is smaller because block is shorter along the depth direction and has a very large cross-sectional area.  Both of these changes decrease the resistance.

5. (moderate) If a conducting wire (X) has a resistance of R_x, determine the resistance of a second wire (Y) if the resistivity is twice the resistivity of X, but is only half as long and has three times the diameter.
The radius of Y is 3 times that of X. Thus, the cross-sectional area of Y is 9 times that of X
R_x = ρ_xL_x/A_x
R_y = 2ρ_x(L_x/2)/9A_x = R_x/9

6. (easy) A resistor is connected to a 20 v emf. The current through the resistor is found to be 2A. What current would move through the resistor if it was connected to a 50 v source? Find the rate at which the energy is dissipated by the resistor with the 50 v source.
Find the resistance:
V = IR
20 = 2R
R = 10Ω
Now find the current when connected to the 50 v source:
V = IR
50 = I(10)
I = 5A 
Find the power using the 50 v source:
P = I²R
P = (5)²(10)
P = 250 watts  

7. (easy) Explain why the sketch shown below is a good analogy for Ohm's Law.

Ohm's Law relates the flow of charge to the electrical resistance and electric potential difference (I = V/R). This sketch relates the flow of water to the clogging (resistance to flow) in the pipe and the pressure caused by the height of the water in the system. The sketch shows that the flow (current) can be the same when the pressure (voltage) is increased when the clogging (resistance) increases. 

8. (easy) Determine the current in a circuit that contains a 10 volt emf and 3 Ohmic resistors in series. Ohmic resistors are resistors that obey Ohm's Law. (Non-Ohmic do not obey Ohm's Law). You will find in the next lecture, that when resistors are connected in series, the total resistance is the simple addition of the individual resistors. The banded resistance on the resistors are as follows:
Resistor #1: Green|Yellow|Red|Gold
Resistor #2: Blue|Red|Brown|Gold
Resistor #3: Red|Red|Red|Silver
Resistor #1 = 5400 Ω
Resistor #2 = 620 Ω
Resistor #3 = 2200 Ω
Total Resistance = 5400 + 620 + 2200 = 8220 Ω
For the circuit:
V = IR
10 = I(8220)
I = 0.0012 A 

9. (moderate) Estimate the cost of cooking a turkey for 4 hours in an oven that operates at 20 A and 240 v. Assume the cost for 1.0 kwh is 8 cents.
P = IV = (20 A)(240v) = 4800 watts = 4.8 kw
Energy = Pt = (4.8 kw)(4h) = 19.2 kwh
cost = (19.2 kwh)($0.08/kwh) = $1.54 

10. (moderate) A heating element in a kitchen appliance uses a voltage of 120 volts. The device dissipates 10000 J of energy in 10 seconds. Find both the resistance of the heating element and the current that is established.
P = E/t = 10000/10 = 1000 watts
P = V²/R
R = V²/P = (120)²/1000 = 14.4 Ω
V = IR
120 = I(14.4)
I = 8.3 A 

11. (moderate) A certain toaster has a heating element made of nichrome wire. When first connected to a 120 v emf the wire is at 20°C and the initial current is 1.8 A. When the toaster reaches its final operating temperature the current in the nichrome wire is 1.53 A. Determine the power consumed at the final temperature and the final temperature itself. Note: The resistivity changes with temperature according to this equation: ρ = ρ_o(1 + 0.0004(T_f - 20)).
(Also, assume that the size of the wire doesn't change due to thermal expansion)
Initially: R_o = V_o/I_o = 120/1.8 = 66.7 Ω
Finally: R_f = V_f/If = 120/1.53 = 78.4 Ω
If the area and length don't appreciably change, the resistivity equation can be used to find the final operating temperature.
R_o = ρ_oL_o/A_o  and ρ_o = R_oA_o/L_o 
R_f = ρ_fL_f/A_f    and ρ_f = R_fA_f/L_f  
ρ = ρ_o(1 + 0.0004(T_f - 20)) 
Thus, R_f = R_o(1 + 0.0004(T_f - 20))
78.4 = 66.7(1 + 0.0004(T_f - 20))
T_f = 459°C

12. (easy) A 3 volt battery is connected in series to two identical resistors (R = 10 Ω). First, determine the current from the battery if the resistors are connected in series and then find the current if the resistors are connected in parallel.
Series connection:
R_eq = 10 + 10 = 20Ω
V = IR
3 = I(20)
I = 0.15 A
Parallel connection:
1/R_eq = 1/10 + 1/10
R_eq = 5Ω
V = IR
3 = I(5)
I = 0.6 A

13. (moderate) Determine the power dissipated by each resistor in problem #12. What is the total power in each case?
P = I²R
For the series connection:
P = (0.15)²(10) =0.23 watts (each resistor has the same power)
Total power = 2(0.23) = 0.46 watts
For the parallel connection:
The total current (from the battery) is 0.6 A, but each resistor gets half that current, 0.3 A.
P = (0.3)²(10) = 0.9 watts (each resistor has the same power)
Total power = 2(0.9) = 1.8 watts

14. Three identical resistors (R) are connected to a battery in a simple circuit. What equivalent resistances can the circuit exhibit?
R/3, 3R/2, 5R/2, R/2

When connected in parallel, the three resistors will have an equivalent resistance of R/3. If two of the resistors are in parallel and those two are in series with the third resistor, the total resistance would be 3R/2. If all three resistors are in series, the total resistance is 3R. But that resistance is not in the list.

15. (moderate) A 10 volt battery, a 60 Ω resistor, a light bulb and a fuse (R = 0.02 Ω) are connected in series. The light bulb dissipates 100 watts when connected across a 20 v source. If the fuse is rated at 0.1 A, will the circuit operate?
Find the resistance of the bulb:
P = V²/R_bulb 100 = (20)²/R_bulb
R_bulb = 4Ω
Find the effective resistance:
R_eq = 60 + 4 + 0.02 = 64.02Ω
Find the current from the battery:
V = IR_eq
10 = I(64.02)
I = 0.16 A
The fuse will burn out and the circuit will not operate

16. (moderate) If the 60 Ω resistor in the previous question was replaced with a larger resistor, what would its minimum resistance be so that the circuit would operate?
Find the minimum value for the total resistance when the current is 0.1 A.
V = IR_eq
10 = (0.1)R_eq
R_eq = 100 Ω
R_eq = R + R_bulb + R_fuse
100 = R + 4 + 0.2
R = 95.8 Ω
(This is the minimum resistance needed)

17. (moderate) Analyze the circuit below in order to fill in the chart.

The 4Ω and 5Ω resistors combine to make a 2.2Ω resistor
The 6Ω, 7Ω and 8Ω resistors combine to make a 2.3Ω resistor
The total resistance in the circuit: R_EQ = 3Ω + 2.2Ω + 2.3Ω = 7.5Ω
The current out of the battery: E = IR
30 = I(7.5)
I = 4.0 A
The voltage drop across the 4Ω and 5Ω resistors:
V = 4.0(2.2) = 8.8 volts
The current through the 4Ω resistor:
8.8 = I(4)
I = 2.2 A
P = IV = 2.2(8.8) = 19.4 watts 
The current through the 5Ω resistor:
8.8 = I(5)
I = 1.8 A
P = IV = 1.8(8.8) = 15.8 watts 
The voltage drop across the 6Ω, 7Ω and 8Ω resistors:
V = 4.9(2.3) = 9.2 volts
The current through the 6Ω resistor:
9.2 = I(6)
I = 1.5A
P = IV = 1.5(9.2) = 13.8 watts 
The current through the 7Ω resistor:
9.2 = I(7)
I = 1.3A
P = IV = 1.3(9.2) = 12.0 watts
The current through the 8Ω resistor:
9.2 = I(8)
I = 1.2A
P = IV = 1.2(9.2) = 11.0 watts

18. A teacher challenges a group of students to create a circuit that costs $1.00 to operate over two days. The cost per kwh is $0.20 in the lab. The students must use 4 identical lightbulbs (not all in series) in the circuit and a 20 v battery.
a. Design a circuit that would allow the students to meet the challenge. You must report the resistance of the lightbulbs you choose to use.
Based on the constraints given in the problem, the power of the circuit can be calculated:
cost = energy(cost/energy)
$1.00 = E($0.20/1 kwh)
Energy = 5 kwh
Power = Energy/time
P = 5 kwh/48 h
P = 0.1 kw = 100 watts
A circuit using 4 bulbs (resistors), each with 1.6 Ω of resistance connected as shown below would cost $1.00 to operate over the 2 day period.
The total resistance would be 4.0 Ω and the power would be calculated as:
P = V²/R
P = 20²/4.0 = 100 watts
Other answers are possible using different bulbs connected in a different fashion. 

b. If all the bulbs determined in part (a) were used in series, would the current from the battery increase or decrease and would the cost to operate the circuit over two days increase or decrease? Explain your answer.
If the bulbs were all in series the total resistance would increase, causing the current to decrease according to Ohms' Law (V = IR), prompting a decrease in power according to Joule's Law (P = V²/R). Since the power decreases so would the cost to operate the device.  

19. (moderate) Find the voltage across and the current through the 330Ω resistor and the middle 560Ω resistor in the circuit shown below.