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 2 of 3.
Clinton said:
4 years ago
V=IR.
R= V/I.
i.e V= 18V.
I= 40ma.
R=18/40= 0.45.
V =IR.
I = 20ma.
R =0.45.
V = 20x0.45 = 9V.
R= V/I.
i.e V= 18V.
I= 40ma.
R=18/40= 0.45.
V =IR.
I = 20ma.
R =0.45.
V = 20x0.45 = 9V.
(6)
Swetha said:
7 years ago
Vg is directly proportional to I.
Ohm's law applies.
V1/V2=I1/I2 substitute this formula....
Ohm's law applies.
V1/V2=I1/I2 substitute this formula....
Krishna said:
9 years ago
Or you can say V is directly proportional to I. So when the current drops so as the voltage.
Ladli said:
7 years ago
Here, the load is constant(R).
So,v/i=R=v'/i'.
18/40mA=v'/20mA.
Hence, v'=9volt.
So,v/i=R=v'/i'.
18/40mA=v'/20mA.
Hence, v'=9volt.
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.
Srk said:
8 years ago
v1=18 i1=40ma,i2=20ma,v2=?
R=18/40ma = 450.
V2 = i2 * r = 20m * 450 = 9a.
R=18/40ma = 450.
V2 = i2 * r = 20m * 450 = 9a.
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
Estrellanes said:
9 years ago
The half of 40 is 20, and the half of 18 is 9. So the answer is A) 9v.
Sunand Behera said:
1 decade ago
Circuit resistance remains constant and then follow as Nikul's answer.
Hardik said:
6 years ago
V = IR.
V directly proportional to I.
So, Apply proportional rules.
V directly proportional to I.
So, Apply proportional rules.
(1)
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