Facebook
Google
GitHub
LinkedinBasic Electricity
Ohm’s Law
16 questions By Tony R. Kuphaldt
-
Question 16 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 = 45 mA, R = 3.0 kΩ; E =
I = 10 kA, R = 0.5 mΩ; E =
E = 45 V, R = 4.7 kΩ; I =
E = 13.8 kV, R = 8.1 kΩ; I =
E = 500 μV, I = 36 nA; R =
E = 14 V, I = 110 A; R =
I = 0.001 A, R = 922 Ω; E =
I = 825 A, R = 15.0 mΩ; E =
E = 1.2 kV, R = 30 MΩ; I =
E = 750 mV, R = 86 Ω; I =
E = 30.0 V, I = 0.0025 A; R =
E = 0.00071 V, I = 3389 A; R =
Reveal answerI = 45 mA, R = 3.0 kΩ; E = 140 V = 1.4 ×102 V
I = 10 kA, R = 0.5 mΩ; E = 5 V = 5 ×100 V
E = 45 V, R = 4.7 kΩ; I = 9.6 mA = 9.6 ×10−3 A
E = 13.8 kV, R = 8.1 kΩ; I = 1.7 A = 1.7 ×100 A
E = 500.0 μV, I = 36 nA; R = 14 kΩ = 1.4 ×104 Ω
E = 14 V, I = 110 A; R = 130 mΩ = 1.3 ×10−1 Ω
I = 0.001 A, R = 922 Ω; E = 900 mV = 9 ×10−1 V
I = 825 A, R = 15.0 mΩ; E = 12.4 V = 1.24 ×101 V
E = 1.2 kV, R = 30 MΩ; I = 40 μA = 4 ×10−5 A
E = 750 mV, R = 86 Ω; I = 8.7 mA = 8.7 ×10−3 A
E = 30.0 V, I = 0.0025 A; R = 12 kΩ = 1.2 ×104 Ω
E = 0.00071 V, I = 3389 A; R = 210 nΩ = 2.1 ×10−7 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.
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?