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.
Suresh babu said:
1 decade ago
V=IR
V=(0.750)*(11)
V=8.25
V=(0.750)*(11)
V=8.25
(1)
Anil said:
1 decade ago
v=i*r
Punithajothi said:
1 decade ago
v=i*r its simple
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)
Vshree said:
1 decade ago
According to Ohm's Law: V = I x R
Joy jumbo itah said:
1 decade ago
Given that;
I=750microamps ie;750/1000 =0.750A
R=11kil0 ohms
therefore V = I*R from ohm's law
V = 0.75*11=8.25v
It's right
I=750microamps ie;750/1000 =0.750A
R=11kil0 ohms
therefore V = I*R from ohm's law
V = 0.75*11=8.25v
It's right
4getfull said:
1 decade ago
Regarding the ohms wouldn't k represent 1,000, if not what is the abbreviation represent? or did i just get it it by realizing the microamps has already been converted by devising by 1,000 and thats why 11 works instead of 11,000
Nju said:
1 decade ago
According to my concept too 11kilo ohms must be 11000ohms.
So do I disagree with the ans i.e 8. 25v.
So do I disagree with the ans i.e 8. 25v.
M.Nedunchezhian said:
1 decade ago
Ya I have same doubt like @4getfull.
M.Nedunchezhian said:
1 decade ago
All above methods are wrong.
1 microamphere = 10^-6 Amphere.
Therefore, 750 microamphere = 750/1000000=0.00075 Amphere.
11 kiloohm = 11000 ohms.
V=I*R
V=0.00075*11000=8.25V.
1 microamphere = 10^-6 Amphere.
Therefore, 750 microamphere = 750/1000000=0.00075 Amphere.
11 kiloohm = 11000 ohms.
V=I*R
V=0.00075*11000=8.25V.
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