1. 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 =
E
R
R =
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:
1. I = 20 mA, R = 5 kΩ; E =
2. I = 150 µA, R = 47 kΩ; E =
3. E = 24 V, R = 3.3 MΩ; I =
4. E = 7.2 kV, R = 900 Ω; I =
5. E = 1.02 mV, I = 40 µA; R =
6. E = 3.5 GV, I = 0.76 kA; R =
7. I = 0.00035 A, R = 5350 Ω; E =
8. I = 1,710,000 A, R = 0.002 Ω; E =
9. E = 477 V, R = 0.00500 Ω; I =
10. E = 0.02 V, R = 992,000 Ω; I =
11. E = 150,000 V, I = 233 A; R =
12. E = 0.0000084 V, I = 0.011 A; R =
13. I = 45 mA, R = 3.0 kΩ; E =
14. I = 10 kA, R = 0.5 mΩ; E =
15. E = 45 V, R = 4.7 kΩ; I =
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.
Solve for Power
1. E = 0.0000084 V, I = 0.011 A; P =
2. I = 45 mA, R = 3.0 kΩ; P=
3. I = 10 kA, R = 0.5 mΩ; P=
4. E = 45 V, R = 4.7 kΩ; P=
5. E = 13.8 kV, R = 8.1 kΩ; P=
6. E = 500 µV, I = 36 nA; P=
7. E = 14 V, I = 110 A; P=
8. I = 0.001 A, R = 922 Ω; P=
9. I = 825 A, R = 15.0 mΩ; P=
10. E = 1.2 kV, R = 30 MΩ; P=
TAKE ADVANTAGE OF OUR PROMOTIONAL DISCOUNT DISPLAYED ON THE WEBSITE AND GET A DISCOUNT FOR YOUR PAPER NOW!
Call Us: USA +1 862 207 3288