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 1 of 3.
PROJIT SAHA said:
1 decade ago
ACCORDING TO OHM'S LAW,
R(resistance)=V(voltage)/I(current)
=12/3.6
=3.33 kilo-ohm
R(resistance)=V(voltage)/I(current)
=12/3.6
=3.33 kilo-ohm
Vinay said:
1 decade ago
Ohm's Law :- V = I*R
But here we adjust the terms as,
R = V/I
= 12/3.6 * 10^-3 ...(^ sign is POWER )
= (12/3.6)* 10^3 ...(shifting of 10^-3)
= 3.3 * 10^3
R = 3.3 kilo-ohm
But here we adjust the terms as,
R = V/I
= 12/3.6 * 10^-3 ...(^ sign is POWER )
= (12/3.6)* 10^3 ...(shifting of 10^-3)
= 3.3 * 10^3
R = 3.3 kilo-ohm
Mani said:
1 decade ago
I want range of mA and killo-ohm. Any one say?
Surekha said:
1 decade ago
mA=10^-3
kilo ohm =10^3
kilo ohm =10^3
Sakthi said:
1 decade ago
V=I/R
V=12/3.6
V=3.3KM
V=12/3.6
V=3.3KM
Anand said:
1 decade ago
Nice answer vinay
Suraj said:
1 decade ago
Absolutely its 3.33kilo ohms.
Tapaswini swain said:
1 decade ago
I=V/R
V=12V
I=3.6mA=3.6*10^-3
R=V/I
R=(12/36*10^-4)Ohm
=(1/3)*10^4ohm
=0.333*10^4ohm
=3.33*10^3ohm
=3.33kilo-ohm
V=12V
I=3.6mA=3.6*10^-3
R=V/I
R=(12/36*10^-4)Ohm
=(1/3)*10^4ohm
=0.333*10^4ohm
=3.33*10^3ohm
=3.33kilo-ohm
Vikas kumar said:
1 decade ago
V=12, I = 3.6mA=0.0036A.
V=IR.
R=V/I=12/0.0036=12*10^4/36=10^4/3=10000/3=3333.33 ohm.
3.3 kilo ohm.
V=IR.
R=V/I=12/0.0036=12*10^4/36=10^4/3=10000/3=3333.33 ohm.
3.3 kilo ohm.
Sunny said:
1 decade ago
V=12
I=3.6mA
R=12/3.6*10^-3=12/0.036=3.3kohms.
I=3.6mA
R=12/3.6*10^-3=12/0.036=3.3kohms.
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