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?
27.2 mA
59 mA
5.9 mA
590 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
29 comments Page 1 of 3.

PROJIT said:   1 decade ago
I=V/R
=40/6.8
=5.88

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?

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.

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

Anand said:   1 decade ago
Nice answer Raut mahesh

Jayesh punekar said:   1 decade ago
I=V/R
I=40/6.8
I=5.88 mA

PRIYA said:   1 decade ago
V=IR
I=V/R
V=40
R=6.8
I=5.9mA

Sunny said:   1 decade ago
v=40v
r=6.8kv=6.8*10^3=6800
i=40/6800=0.059A=5.9A

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.

Pooja said:   1 decade ago
Given that,

V = 40v,
R = 6.8Kohm,
I = V/R.

I = 40/6.8*(10^-3).

I = 5.882mA.

That is right answer.


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