Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 18)
18.
A current of 200 µA through a 6.8 k
resistor produces a voltage drop of
resistor produces a voltage drop ofDiscussion:
7 comments Page 1 of 1.
Bhaveshkumar B Detroja said:
10 years ago
V : I*R.
V : 200*10^-6*6.8*10^3.
V : 200*10^3*6.8.
V : 0.2*6.8.
V : 1.36 VOLTS.
V : 200*10^-6*6.8*10^3.
V : 200*10^3*6.8.
V : 0.2*6.8.
V : 1.36 VOLTS.
Babu said:
1 decade ago
200/100000*6.8*1000 = 13.6 v.
(1)
Shravan said:
1 decade ago
V = IR.
First covert.
200*10^-6 = 0.0002 A.
6.8*10^3 = 6800 ohm.
0.0002*6800 = 1.36 v.
First covert.
200*10^-6 = 0.0002 A.
6.8*10^3 = 6800 ohm.
0.0002*6800 = 1.36 v.
Unknown said:
1 decade ago
V = IR.
V = 200x0.0068.
V = 1.36.
V = 200x0.0068.
V = 1.36.
Pankaj said:
1 decade ago
I did not understand this answer.
Bhagirath said:
1 decade ago
V=IR
=(200*6.8)*1000/1000000
=200*6.8/1000
=1360/1000
=1.360
=(200*6.8)*1000/1000000
=200*6.8/1000
=1360/1000
=1.360
Rocky said:
1 decade ago
According to Ohm's law, V=IR,
therefore ((200x10^-6)x(6.8x10^3))=1.36v
therefore ((200x10^-6)x(6.8x10^3))=1.36v
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