Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 2)
2.
If 750 µA is flowing through 11 k
of resistance, what is the voltage drop across the resistor?
of resistance, what is the voltage drop across the resistor?Discussion:
60 comments Page 5 of 6.
VINCENT said:
5 years ago
V = IR.
V = (750/1000)(11),
ASW = 8.25 V.
V = (750/1000)(11),
ASW = 8.25 V.
(5)
Madhuri Gulhane said:
8 years ago
V = IR.
V = 750*10^-6*11*10^3.
V = 8.25V.
V = 750*10^-6*11*10^3.
V = 8.25V.
ABHINANDAN Asd said:
8 years ago
Thanks for your solution @Nedunchezhian.
Sanjeeb Kumar Behera said:
7 years ago
V = IR
V = (0.750)*(11)
V = 8.25. (A).
V = (0.750)*(11)
V = 8.25. (A).
BHARAT said:
1 decade ago
V = IR.
V = 0.000750X11000.
V = 8.25 V.
V = 0.000750X11000.
V = 8.25 V.
Johnlerr espenoza said:
2 years ago
V = R×I.
V = 11,000 × 750.
= 8.25V.
V = 11,000 × 750.
= 8.25V.
(9)
M.Nedunchezhian said:
1 decade ago
Ya I have same doubt like @4getfull.
Sai said:
4 years ago
750*10^-6 = 0.00075 * 11000 = 8.25.
(5)
Vshree said:
1 decade ago
According to Ohm's Law: V = I x R
Samir said:
1 decade ago
Simple V = IR.
0.75*11 = 8.25.
0.75*11 = 8.25.
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