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 3 of 3.
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.
Sunil Pant said:
1 decade ago
40mA -> 18V
Then 20mA -> (?)
20*18/40 = 9V
Then 20mA -> (?)
20*18/40 = 9V
MATHAN SENTHIL said:
1 decade ago
(V1/I1)=(V2/I2)
THEREFORE V2=(V1*I2)/I1
V2=(18*20*10^-3)/(40*10^-3)= 9V
THEREFORE V2=(V1*I2)/I1
V2=(18*20*10^-3)/(40*10^-3)= 9V
ASHOK said:
1 decade ago
V=IR(OHM'S LAW)
R=V/I
TAKE RATIO:V1/I1=V2/I2
V2=(V1*I2)/I1
V2=(18*20)/40
V2=9V.
R=V/I
TAKE RATIO:V1/I1=V2/I2
V2=(V1*I2)/I1
V2=(18*20)/40
V2=9V.
Kavitha said:
1 decade ago
V1/I1=V2/I2
V2=(V1*I2)/I1
V2=(18*20)/40=9V
V2=(V1*I2)/I1
V2=(18*20)/40=9V
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).
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
Sunand Behera said:
1 decade ago
Circuit resistance remains constant and then follow as Nikul's answer.
Nikul Prajapati said:
1 decade ago
40mA - 18V
then 20mA - (?)
20*18/40 = 9V
then 20mA - (?)
20*18/40 = 9V
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