Seed = 3324. (If you are not the instructor, you can ignore the seed.)
**** SOLUTION(S) ****



Problem 1. A light bulb passes 515 mA of current when 139 V is applied.
a. What is the bulb's effective resistance?
b. How much power is the bulb using?
R = V / I = (139 V) / (0.515 A) = 270 Ω
P = V * I = (139 V) * (0.515 A) = 71.6 W


Problem 2. If an electric company charges 40.7 cents/kWh, calculate the cost for a 85.6 W bulb if it's left on for 3.77e+03 hours.
a) 20400 dollars
b) 17000 dollars
c) 196 dollars
d) 131 dollars
e) None of these

The correct answer is d) 131 dollars



Problem 3. Two resistors, R1 = 300 Ω and R2 = 696 Ω are used in conjunction with a 14.8 V power supply.
a. If the resistors are placed in series, what is the current through, and the voltage across, each resistor?
b. If the resistors are placed in parallel, what is the current through, and the voltage across, each resistor?
a.
RT = 996 Ω.
IT = VT / RT = (14.8 V) / (996 Ω) = 0.0149 A, is the same for each resistor.
V1 = IT R1 = 4.47 V
V2 = 14.8 V - 4.47 V = 10.3 V


b.
The voltage across each resistor is the same as the total voltage, i.e. 14.8 V.
RT = R1 * R2 / (R1 + R2) = 210 Ω.
IT = VT / RT = (14.8 V) / (210 Ω) = 0.0705 A
I1 = VT / R1 = (14.8 V) / (300 Ω) = 0.0493 A.
I2 = IT - I1 = 0.0212 A




Problem 4. Two resistors, RA and RB, are placed in parallel. If RA > RB and a voltage is applied across this resistor combination, which resistor will dissipate more power?
a) RB
b) Impossible to determine; need more information to answer
c) RA
d) They will dissipate the same amount of power

The correct answer is a) RB



Problem 5. In the figure below, let VS = 11.8 V, R1 = 17.3 Ω, R2 = 29.1 Ω, and R3 = 18.8 Ω.






a. What is the current through R3?
b. What is the power dissipated by R2?

a. The current through R3 is the same as the total current, IT.
RT = R3 + (1/R1 + 1/R2)-1 = 18.8 + (10.8) = 29.6 Ω.
IT = VS / RT = 11.8 / 29.6 = 0.399 A


b. Two ways to work this:
V2 = VS - IT * R3 = 11.8 - (0.399 A)(18.8 Ω) = 4.3 V
P2 = V2² / R2 = 0.635 W

or, using the current divider formula
I2 = IT R12 / R2 = 0.148 A
P2 = I2² R2 = 0.637 W, within roundoff error of 0.635 W



Problem 6. Shown below is a circuit for providing a variety of output voltages using a switch, labeled S1.
Find the current through the load resistor, R_L, for both switch settings.
Let V_S = 99.8 V, R_A= 122 Ω, R_B = 169 Ω and R_L = 108 Ω.






S1 open:
R_T = R_A+ R_B + R_L = 122 + 169 + 108 = 399 Ω.
I_T = V_S / R_T = 99.8 / 399 = 0.300 A


S1 closed:

IT = V_S / R_T = 99.8 / 207 = 0.482 A
V_LB = V_S - I_T R_A = 99.8 - 0.482 (122) = 41 V
I_LB = (41 V) / (277 Ω) = 0.148 A