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 1 of 6.
Johnlerr espenoza said:
2 years ago
V = R×I.
V = 11,000 × 750.
= 8.25V.
V = 11,000 × 750.
= 8.25V.
(9)
Kranthi bk said:
2 years ago
Thanks all for explaining.
(10)
Lakshmi prasanna said:
3 years ago
Thanks for giving the clear explanation @M. Neduchezhian.
(3)
Sai said:
4 years ago
750*10^-6 = 0.00075 * 11000 = 8.25.
(5)
Mohammed Faizoddin said:
5 years ago
According to ohms law, I=750*10^-6
V=0.00075*11000
=8.25v.
V=0.00075*11000
=8.25v.
(2)
M. Rama krishna said:
5 years ago
I=750*10^-6.
R=11*10^3.
V=0.75*11,
= 8.25v.
R=11*10^3.
V=0.75*11,
= 8.25v.
(2)
VINCENT said:
5 years ago
V = IR.
V = (750/1000)(11),
ASW = 8.25 V.
V = (750/1000)(11),
ASW = 8.25 V.
(5)
Haroon Zaman said:
5 years ago
V = IR.
V = (750*10^-6)(11*10^3),
V = 8250*10^-3,
V = 8.25*10^3*10^-3,
V = 8.25.
V = (750*10^-6)(11*10^3),
V = 8250*10^-3,
V = 8.25*10^3*10^-3,
V = 8.25.
(4)
Vishal said:
5 years ago
Good Explanation, thanks all.
Rahul Singh Rathore said:
5 years ago
I=750mic.amp =.750mili amp =.000750 amp.
R= 11 k ohms = 11000 ohm,
V= IR=. 000750 * 11000 =.750 * 11 = 8.25V.
R= 11 k ohms = 11000 ohm,
V= IR=. 000750 * 11000 =.750 * 11 = 8.25V.
(9)
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