# scientific notation

scientific notation

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 ()

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 k; E =

2. I = 150 µA, R = 47 k; E =

3. E = 24 V, R = 3.3 M; I =

4. E = 7.2 kV, R = 900 ; 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 ; E =

8. I = 1,710,000 A, R = 0.002 ; E =

9. E = 477 V, R = 0.00500 ; I =

10. E = 0.02 V, R = 992,000 ; 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 k; E =

14. I = 10 kA, R = 0.5 m; E =

15. E = 45 V, R = 4.7 k; 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 M). 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 k; P=

3. I = 10 kA, R = 0.5 m; P=

4. E = 45 V, R = 4.7 k; P=

5. E = 13.8 kV, R = 8.1 k; P=

6. E = 500 µV, I = 36 nA; P=

7. E = 14 V, I = 110 A; P=

8. I = 0.001 A, R = 922 ; P=

9. I = 825 A, R = 15.0 m; P=

10. E = 1.2 kV, R = 30 M; P=

TAKE ADVANTAGE OF OUR PROMOTIONAL DISCOUNT DISPLAYED ON THE WEBSITE AND GET A DISCOUNT FOR YOUR PAPER NOW!