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 2 of 6.
Benard Istifanus said:
6 years ago
Thanks all for explaining it.
Aasai thambi said:
6 years ago
i=750 microamps =>0.000750 amps.
r=11000ohms,
0.000750*11000 =>0.750 * 11=>8.25v.
r=11000ohms,
0.000750*11000 =>0.750 * 11=>8.25v.
Meghray said:
7 years ago
Ans is [B] 82.5.
Given data: I = 750microAmps
= 750 * 10^-6 amps.
Resistance(R)= 11kohm= 11*10^3ohm
Voltage drop (Vd)= I*R.
= (750 *10^-6 )*(11*10^3),
= 82.5V.
Given data: I = 750microAmps
= 750 * 10^-6 amps.
Resistance(R)= 11kohm= 11*10^3ohm
Voltage drop (Vd)= I*R.
= (750 *10^-6 )*(11*10^3),
= 82.5V.
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).
Kadija said:
7 years ago
Micro amp is different from milli amp.
I = in Amp.
Amp -> millli Amp -> micro Amp,
1 --> 1000 --> 1000000.
So, I thought we should divide 750 microamps into 1000000 to get the unit Amp.
Please clarify.
I = in Amp.
Amp -> millli Amp -> micro Amp,
1 --> 1000 --> 1000000.
So, I thought we should divide 750 microamps into 1000000 to get the unit Amp.
Please clarify.
Faisal said:
7 years ago
8.25 is the correct answer.
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.
Braj Kumar said:
8 years ago
Right answer is 8.25 volt.
Indranil das said:
8 years ago
voltage drop refers to the voltage consumed by the particular resistance,
Here it is 8.25v and supply voltage whatever may be (here it is not mentioned) should be greater than 8.25v.
Here it is 8.25v and supply voltage whatever may be (here it is not mentioned) should be greater than 8.25v.
JANA333 said:
8 years ago
How to calculate micro amps to amps?
1A=10^3 mA
1A=10^6
V=IR,
I=( 750/10^6),
=0.00075A THEN,
11K OHM= (11*1000),
= 11000 OHMS,
V=(11000*0.00075) =8.25V Ans.
I think everyone understands this
1A=10^3 mA
1A=10^6
V=IR,
I=( 750/10^6),
=0.00075A THEN,
11K OHM= (11*1000),
= 11000 OHMS,
V=(11000*0.00075) =8.25V Ans.
I think everyone understands this
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