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?
8.25 V
82.5 V
14.6 V
146 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
60 comments Page 3 of 6.

Khan said:   1 decade ago
What is the difference between actual supply & voltage drop.

Samir said:   1 decade ago
Simple V = IR.

0.75*11 = 8.25.

Srivani said:   1 decade ago
Current(I)=750X10^-6 A,

R=11X10^3 ,

V=I*R=750X10^-6*11X10^-3=750X11X(10^-6+10^3).

V=750X11X10^-3.

V=0.75X11=8.25 V.

Deepak mp said:   1 decade ago
V = IR.
V = 750*11 = 8250.
V = 8250.

V/1000.
8250/1000 = 8.25.

Ans = 8.25V.

M.Arasumani said:   1 decade ago
V = IR.
V = ?

I = 750x10^-6(micro amp to amp).
R = 11000(kilo ohms to ohms).

First all the unit come same unit ohms, amp, volt that's way answer correct other wise wrong.

V = 8.25.

TUSHAR K NASKAR said:   1 decade ago
V = IR.
V = 750*11.
V = 8250.
V = 8250/1000.
V = 8.250 V.

Suman Lal Bharti said:   1 decade ago
Given that,

Current, I= 750 Micro-Ampere= 750/1000000 Amp.= .00075 Amp.
Resistance, R= 11 Kilo-Ohms= 11*1000 Ohms= 11000 Ohms.

Now by using Ohm's Law,
V=I*R;

Hence,
V= .00075*11000= 8.25 Volts.

Richa said:   1 decade ago
1 Micro amp = 10^-6 Amp.
1 Kilo Ohm = 10^3 ohm.
V = IR.
V = (11*10^3)*(750*10^-6).
= 8.25.

Amar bhuan 'binka' said:   1 decade ago
I = 750 micro amp.
= 0.00075 amp.
= 75/1000000 amp.

R = 11 kilo ohm.
= 11*1000 ohm.
= 110000 ohm.

V = IR.
= 75/1000000*110000.
= 8.25 VOLT ANS.

BHARAT said:   1 decade ago
V = IR.
V = 0.000750X11000.
V = 8.25 V.


Post your comments here:

Your comments will be displayed after verification.