Facebook
Google
GitHub
LinkedinBasic Electricity
Ohm’s Law
16 questions By Tony R. Kuphaldt
-
Question 13 of 16
One of the fundamental equations used in electricity and electronics is Ohm’s Law: the relationship between voltage (E or V, measured in units of volts), current (I, measured in units of amperes), and resistance (R, measured in units of ohms):
$$E=IR \ \ \ \ \ \ \ \ \ \ \ I=\frac{E}{R} \ \ \ \ \ \ \ \ \ \ \ R=\frac{E}{I}$$
Where,
E = Voltage in units of volts (V)
I = Current in units of amps (A)
R = Resistance in units of ohms (Ω)
Solve for the unknown quantity (E, I, or R) given the other two, and express your answer in both scientific and metric notations:
I = 20 mA, R = 5 kΩ; E =
I = 150 μA, R = 47 kΩ; E =
E = 24 V, R = 3.3 MΩ; I =
E = 7.2 kV, R = 900 Ω; I =
E = 1.02 mV, I = 40 μA; R =
E = 3.5 GV, I = 0.76 kA; R =
I = 0.00035 A, R = 5350 Ω; E =
I = 1,710,000 A, R = 0.002 Ω; E =
E = 477 V, R = 0.00500 Ω; I =
E = 0.02 V, R = 992,000 Ω; I =
E = 150,000 V, I = 233 A; R =
E = 0.0000084 V, I = 0.011 A; R =
Reveal answerI = 20 mA, R = 5 kΩ; E = 100 V = 1 ×102 V
I = 150 μA, R = 47 kΩ; E = 7.1 V = 7.1 ×100 V
E = 24 V, R = 3.3 MΩ; I = 7.3 μA = 7.3 ×10−6 A
E = 7.2 kV, R = 900 Ω; I = 8.0 A = 8.0 ×100 A
E = 1.02 mV, I = 40 μA; R = 26 Ω = 2.6 ×101 Ω
E = 3.5 GV, I = 0.76 kA; R = 4.6 MΩ = 4.6 ×106 Ω
I = 0.00035 A, R = 5350 Ω; E = 1.9 V = 1.9 ×100 V
I = 1,710,000 A, R = 0.002 Ω; E = 3.42 kV = 3.42 ×103 V
E = 477 V, R = 0.00500 Ω; I = 95.4 kA = 9.54 ×104 A
E = 0.02 V, R = 992,000 Ω; I = 20 nA = 2 ×10−8 A
E = 150,000 V, I = 233 A; R = 640 Ω = 6.4 ×102 Ω
E = 0.0000084 V, I = 0.011 A; R = 760 μΩ = 7.6 ×10−4 Ω
Notes:In calculating the answers, I held to proper numbers of significant digits. This question is little more than drill for students learning how to express quantities in scientific and metric notations.
-
Question 14 of 16
A quantity often useful in electric circuit analysis is conductance, defined as the reciprocal of resistance:
$$G= \frac{1}{R}$$
The unit of conductance is the siemens, symbolized by the capital letter “S”. Convert the following resistance values into conductance values, expressing your answers in both scientific and metric notations:
R = 5 k Ω ; G =
R = 47 Ω ; G =
R = 500 M Ω ; G =
R = 18.2 μΩ ; G =
Now, algebraically manipulate the given equation to solve for R in terms of G, then use this new equation to work “backward” through the above calculations to see if you arrive at the original values of R starting with your previously calculated values of G.
Reveal answerR = 5 k Ω ; G = 200 μS = 2 ×10−4 S
R = 47 Ω ; G = 21 mS = 2.1 ×10−2 S
R = 500 M Ω ; G = 2 nS = 2 ×10−9 S
R = 18.2 μΩ ; G = 55 kS = 5.5 ×104 S
Solving for R in terms of G:
$$R= \frac{1}{G}$$
Notes:Ask your students to show you exactly how they manipulated the equation to solve for R. The last instruction given in the question - working backward through the five calculations to see if you get the original (given) resistance values in ohms - is actually a very useful way for students to check their algebraic work. Be sure to make note of this in class!
-
Question 15 of 16
Suppose an electric current of 1.5 microamps (1.5 μA) were to go through a resistance of 2.3 mega-ohms (2.3 MΩ). How much voltage would be “dropped” across this resistance? Show your work in calculating the answer.
Reveal answer1.5 ×10−6 amps of current through a resistance of 2.3 ×106 Ω will produce a voltage “drop” equal to 3.45 volts.
Notes:It is important for students to understand that metric prefixes are nothing more than “shorthand” forms of scientific notation, with each prefix corresponding to a specific power-of-ten.
The notes for question 14 say “working backwards through the five calculations to see if you get the original (given) figures in degrees Celsius”. Why is Celsius mentioned?