Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 6)
6.
How much resistance is required to limit the current from a 12 V battery to 3.6 mA?
3.3 k
33 k
2.2 k
22 k
Discussion:
24 comments Page 3 of 3.
Densy said:
6 years ago
Exa (E) =10^18.
Peta(P)=10^15.
Tera(T)=10^12,
Giga(G)=10^9,
Mega(M)=10^6,
Kilo(K)=10^3,
Hecto(h)=10^2,
Deka(da)=10^1,
Deci(d)=10^-1,
Centi(c)=10^-2,
Milli(m)=10^-3,
Micro(mue)=10^-6,
Nano(n)=10^-9,
Pico(p)=10^-12,
Femto(f)=10^-15,
Atto(a) = 10^-18.
Peta(P)=10^15.
Tera(T)=10^12,
Giga(G)=10^9,
Mega(M)=10^6,
Kilo(K)=10^3,
Hecto(h)=10^2,
Deka(da)=10^1,
Deci(d)=10^-1,
Centi(c)=10^-2,
Milli(m)=10^-3,
Micro(mue)=10^-6,
Nano(n)=10^-9,
Pico(p)=10^-12,
Femto(f)=10^-15,
Atto(a) = 10^-18.
(7)
Anil said:
4 years ago
According to Ohms law;
R = V/I.
R = 12/3.6 = 3.33.
R = V/I.
R = 12/3.6 = 3.33.
(2)
Aakarsh Chandra said:
3 years ago
V = IR ,
R = V/I,
R = 12/0.36A,
R = 33.3 ohm.
R = V/I,
R = 12/0.36A,
R = 33.3 ohm.
(2)
Johnlerr Espenoza said:
2 years ago
R = V÷I.
R = 12÷3.6.
= 3.3Ohms.
R = 12÷3.6.
= 3.3Ohms.
(3)
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