Question

problems
1 express each of the following quantities using engineering notation
a) 12 x 10³ a=

b) 0.000045 a=

c) 250,000,000 ω=

d) 36 x 10⁻² a=

2 convert the following electrical quantities to a decimal value
a) 20 ma=

b) 1.5kω=

c) 10mω=

d) 120 kv=

Explanation:

Step1: Recall engineering - notation rules

Engineering notation is a version of scientific - notation where the exponent of 10 is a multiple of 3.

Step2: Solve part 1a

$12\times10^{3}\text{ A}=12\text{ kA}$ (since $10^{3}$ is kilo - prefix).

Step3: Solve part 1b

$0.000045\text{ A}=45\times10^{- 6}\text{ A}=45\ \mu\text{A}$ (since $10^{-6}$ is micro - prefix).

Step4: Solve part 1c

$250000000\ \Omega=250\times10^{6}\ \Omega = 250\text{ M}\Omega$ (since $10^{6}$ is mega - prefix).

Step5: Solve part 1d

$36\times10^{-2}\text{ A}=0.36\text{ A}$ (first simplify the scientific notation). In engineering notation, $0.36\text{ A}=360\times10^{-3}\text{ A}=360\text{ mA}$ (since $10^{-3}$ is milli - prefix).

Step6: Recall metric - prefix to decimal conversion rules

$1\text{ mA}=10^{-3}\text{ A}$, $1\text{ k}\Omega = 10^{3}\Omega$, $1\text{ M}\Omega=10^{6}\Omega$, $1\text{ kV}=10^{3}\text{ V}$.

Step7: Solve part 2a

$20\text{ mA}=20\times10^{-3}\text{ A}=0.02\text{ A}$

Step8: Solve part 2b

$1.5\text{ k}\Omega=1.5\times10^{3}\Omega = 1500\Omega$

Step9: Solve part 2c

$10\text{ M}\Omega=10\times10^{6}\Omega=10000000\Omega$

Step10: Solve part 2d

$120\text{ kV}=120\times10^{3}\text{ V}=120000\text{ V}$

Answer:

1a. $12\text{ kA}$
1b. $45\ \mu\text{A}$
1c. $250\text{ M}\Omega$
1d. $360\text{ mA}$
2a. $0.02\text{ A}$
2b. $1500\Omega$
2c. $10000000\Omega$
2d. $120000\text{ V}$