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Electrical Engineering - Ohm's Law - Discussion

@ : Home > Electrical Engineering > Ohm's Law > General Questions - Discussion

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"Everything you can imagine is real."
- Pablo Picasso
5. 

Approximately how many milliamperes of current flow through a circuit with a 40 V source and 6.8 k of resistance?

[A]. 27.2 mA
[B]. 59 mA
[C]. 5.9 mA
[D]. 590 mA

Answer: Option C

Explanation:

No answer description available for this question.


Projit said: (Thu, Jan 20, 2011 01:27:39 PM)    
 
I=V/R
=40/6.8
=5.88

Valli said: (Sat, Feb 12, 2011 08:25:13 AM)    
 
The resistance value is only represent as in ohm. Then how the representation here will be ohm in the calculation?

Vinay said: (Wed, Feb 23, 2011 06:03:09 AM)    
 
To solve this question, we must covert the kilo ohm resistance in ohm resistance & then solve it whatever is the answer convert it into the milliampere.

Raut Mahesh said: (Tue, Jun 28, 2011 05:32:44 AM)    
 
I=V/R
I=40/(6.8*10^3)
I=5.88*10^-3 A
OR
I=5.88 mA

Anand said: (Fri, Jul 22, 2011 07:03:02 PM)    
 
Nice answer Raut mahesh

Jayesh Punekar said: (Sun, Jul 24, 2011 03:33:48 PM)    
 
I=V/R
I=40/6.8
I=5.88 mA

Priya said: (Fri, Jun 1, 2012 03:14:37 PM)    
 
V=IR
I=V/R
V=40
R=6.8
I=5.9mA

Sunny said: (Mon, Aug 27, 2012 07:41:08 PM)    
 
v=40v
r=6.8kv=6.8*10^3=6800
i=40/6800=0.059A=5.9A

Prabhat said: (Fri, Oct 12, 2012 10:27:12 AM)    
 
V=40v.

R=6.8kv=6.8*1000=6800 ohm.

I=40/6800=0.00588A.

So 0.00588A=0.00588*1000=5.88 or 5.9mA.

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