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 4 of 6.
Jagdish Raval said:
1 decade ago
What is the Difference between actual Voltage supply & Voltage Drop?
Sareeran said:
1 decade ago
I=750microamps=0.75A
R=11kilo ohms
V=I*R
Therefore
V=0.75*11
V=8.25
R=11kilo ohms
V=I*R
Therefore
V=0.75*11
V=8.25
(1)
Khan said:
1 decade ago
What is the difference between actual supply & voltage drop.
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)
TUSHAR K NASKAR said:
1 decade ago
V = IR.
V = 750*11.
V = 8250.
V = 8250/1000.
V = 8.250 V.
V = 750*11.
V = 8250.
V = 8250/1000.
V = 8.250 V.
Lakshmi prasanna said:
3 years ago
Thanks for giving the clear explanation @M. Neduchezhian.
(3)
Prabu said:
1 decade ago
Mr.M.Neduchezhian vry crt & nice explanation.
Cyko said:
1 decade ago
@ M. Nedunchezhian very good explained thank you.
Gobimohanraj said:
1 decade ago
Nice and Correct Explanation Mr.M.Neduchezhian
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)
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