Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 12)
12.
You are measuring the current in a circuit that is operated on an 18 V battery. The ammeter reads 40 mA. Later you notice the current has dropped to 20 mA. How much has the voltage changed?
Discussion:
29 comments Page 1 of 3.
Raji said:
10 years ago
Current changed but the resistance remains constant. Hence find resistance from given data i.e. v=18V, i=40mA using the formula V=IR.
Now find voltage for new current (20mA) flowing through same resistor using same formula.
Now find voltage for new current (20mA) flowing through same resistor using same formula.
Ashish said:
1 decade ago
Here it is not clear that weather variable or constant load is connected. Say earlier load which takes 40ma is reduced i.e. load is cut, but voltage is constant, so my view we can also say that voltage are remain same.
Aakarsh Chandra said:
3 years ago
V = I * R,
R = V/I.
R = 18/40*10^-3,
R = 18*1000/40,
R = 18000/40,
R = 450.
And now we Will the voltage dropps.
V = I*R.
V= 20*10^-3 * 450,
V= 20*450/1000,
V = 9000/1000,
V = 9v voltage dropped.
R = V/I.
R = 18/40*10^-3,
R = 18*1000/40,
R = 18000/40,
R = 450.
And now we Will the voltage dropps.
V = I*R.
V= 20*10^-3 * 450,
V= 20*450/1000,
V = 9000/1000,
V = 9v voltage dropped.
(6)
Darshanmaths said:
1 decade ago
V1 = 18V I1 = 40mA.
V2 = ? I2 = 20mA.
As per ohm's law,
As resistance remains constant.
V1/I1 = V2/I2.
Hence substituting,
18/40 = V2/20.
Hence,
V2=0.45*20 = 9V.
Hence V2 = 9V.
V2 = ? I2 = 20mA.
As per ohm's law,
As resistance remains constant.
V1/I1 = V2/I2.
Hence substituting,
18/40 = V2/20.
Hence,
V2=0.45*20 = 9V.
Hence V2 = 9V.
HARDIK said:
10 years ago
V = 18 v, I = 40 mA, R = ?
V = IR.
R = V/I.
= 18/40*10^-3.
= (18/40)*10^3.
= 18000/40.
= 450 ohm.
Then, for V=?, R= 450, I=20 mA.
V = IR.
= (20*10^-3)*450.
= 9000*10^-3.
v = 9V.
V = IR.
R = V/I.
= 18/40*10^-3.
= (18/40)*10^3.
= 18000/40.
= 450 ohm.
Then, for V=?, R= 450, I=20 mA.
V = IR.
= (20*10^-3)*450.
= 9000*10^-3.
v = 9V.
Praveen2kanna said:
1 decade ago
V=I*R;==> R=V/I;
V=18V
INITIALLY CURRENT IS 40mA
SO RESISTANCE = 18/(40*10^-3)=450 ohm
THEN CURRENT DROPS TO 20mA
AS RESISTANCE REMAIN SAME 450 ohm
V=(20*10^-3)*450 = 9V
V=18V
INITIALLY CURRENT IS 40mA
SO RESISTANCE = 18/(40*10^-3)=450 ohm
THEN CURRENT DROPS TO 20mA
AS RESISTANCE REMAIN SAME 450 ohm
V=(20*10^-3)*450 = 9V
ROSHAN said:
9 years ago
Becuase ohms law online define a material body, not a practical area. So if voltage increasing it not necessary to increases current and vice versa.
Azeez said:
6 years ago
V=IR.
i.e. V is directly proportional to I.
Current is changing from 40 mA to 20 mA.
V1/V2= I1/I2.
18/V2=40mA/20mA.
V2=9V.
i.e. V is directly proportional to I.
Current is changing from 40 mA to 20 mA.
V1/V2= I1/I2.
18/V2=40mA/20mA.
V2=9V.
(7)
Eeebhola said:
1 decade ago
We know that from ohms law V=IR
or we can say V directly proportional to current. Thats why answer (a).
or we can say V directly proportional to current. Thats why answer (a).
Pravi said:
1 decade ago
First find the resistance,
R = 18/(40*10^-3) = 450.
V' = 450*20*10^-3.
Then, V' = 0.45*20.
V' = 9 v.
R = 18/(40*10^-3) = 450.
V' = 450*20*10^-3.
Then, V' = 0.45*20.
V' = 9 v.
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