Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 16)
16.
How much resistance is needed to draw 17.6 mA from a 12 volt source?
212 
6.8 k
68
680
Discussion:
10 comments Page 1 of 1.
Vatsal Patel said:
4 years ago
Nice Explanation. Thanks everyone.
(1)
Mr.j said:
6 years ago
@Bhaveshkumar B Detroja nice explanation.
(1)
Victor obanero said:
8 years ago
Convert from milliampere to ampere i.e, 1mA=0.001A or 1A=1000mA.
Meenakshi said:
8 years ago
V=IR , R=12/17.6=0.681.
Then r=680.
Then r=680.
(1)
Bhaveshkumar B Detroja said:
10 years ago
Simple Trick:
R : V/I.
R : 12/17.6 * 10^-3.
R : 12 * 10^3/17.6.
R : 12000/17.6.
R : 120000/176.
R : 681 OHM.
R : V/I.
R : 12/17.6 * 10^-3.
R : 12 * 10^3/17.6.
R : 12000/17.6.
R : 120000/176.
R : 681 OHM.
(1)
Mbacchus said:
1 decade ago
My answer was 681.8 how is that 608? answer D?
ASHISH said:
1 decade ago
I = 17.6mA = 17.6*.001 = .0176 A.
V = 12 volt.
R = V/I => 12/.0176 = 681.81 ohm.
V = 12 volt.
R = V/I => 12/.0176 = 681.81 ohm.
Elisha said:
1 decade ago
v = 12, I = 17.6.
v = IR.
R = 12/17.6.
R = 0.681818181*10^3.
R = 681.
v = IR.
R = 12/17.6.
R = 0.681818181*10^3.
R = 681.
HEMU said:
1 decade ago
R = V/I.
R = 12/17.6.
R = 0.681818181^1000.
R = 681.
R = 12/17.6.
R = 0.681818181^1000.
R = 681.
Anand said:
1 decade ago
V=12,I=17.6X10^-3 A, V=IR,
R=v/I,
R=12/17.6X10^-3 =0.681X10^3=App.680 ohm
R=v/I,
R=12/17.6X10^-3 =0.681X10^3=App.680 ohm
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers