Electronics - Quantities and Units - Discussion

15. 

Add 21 mA and 8000 mu.gifA and express the result in milliamperes.

[A]. 21.8 mA
[B]. 218 mA
[C]. 29 mA
[D]. 290 mA

Answer: Option C

Explanation:

No answer description available for this question.

Parimala said: (Mar 2, 2012)  
21mA + 8mA = 29mA

Savitha U R said: (Apr 17, 2012)  
21*10^-3+8000*10^-6=21*10^-3+8*1000*10^-6
=21*10^-3+8*10^3*(10^-3*10^-3)
=21*10^-3+8*(10^3-3-3)
=21*10^-3+8*10^-3
=(21+8)10^-3
=29*10^-3
=29mA

Vijay Kumar said: (May 25, 2012)  
Since 1micro=10ka power-6.
8000microamp= (8000*10ka power-6) amp.
Again 1amp= (10ka power3) milliamp.
After solving we get 8 milliamp.
Final result= (21+8) milliamp.
=29 milliamp.

Subhani said: (Oct 25, 2012)  
21*10^-3=21mA.
8000*10^-6=8000*10^-3 mA.
Now we can write.
8*10^3*10^-3mA=8mA.
Now;.
21mA+8mA=29mA.

Dedipya said: (Sep 13, 2015)  
= 21*10^-3+8*10^3*10^-6.
= 21*10^-3+8*10^-3.
= 10^-3(21+8) = 29 mA.

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