Electrical Engineering - Ohm's Law - Discussion
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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.