Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 5)
5.
Approximately how many milliamperes of current flow through a circuit with a 40 V source and 6.8 k
of resistance?
of resistance?Discussion:
29 comments Page 3 of 3.
Prabhat said:
1 decade ago
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.
R=6.8kv=6.8*1000=6800 ohm.
I=40/6800=0.00588A.
So 0.00588A=0.00588*1000=5.88 or 5.9mA.
Sunny said:
1 decade ago
v=40v
r=6.8kv=6.8*10^3=6800
i=40/6800=0.059A=5.9A
r=6.8kv=6.8*10^3=6800
i=40/6800=0.059A=5.9A
PRIYA said:
1 decade ago
V=IR
I=V/R
V=40
R=6.8
I=5.9mA
I=V/R
V=40
R=6.8
I=5.9mA
Jayesh punekar said:
1 decade ago
I=V/R
I=40/6.8
I=5.88 mA
I=40/6.8
I=5.88 mA
Anand said:
1 decade ago
Nice answer Raut mahesh
Raut Mahesh said:
1 decade ago
I=V/R
I=40/(6.8*10^3)
I=5.88*10^-3 A
OR
I=5.88 mA
I=40/(6.8*10^3)
I=5.88*10^-3 A
OR
I=5.88 mA
Vinay said:
1 decade ago
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.
Valli said:
1 decade ago
The resistance value is only represent as in ohm. Then how the representation here will be ohm in the calculation?
PROJIT said:
1 decade ago
I=V/R
=40/6.8
=5.88
=40/6.8
=5.88
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