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# Electrical Engineering - Ohm's Law - 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

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.