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?
9 V
900 mV
0 V
18 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
29 comments Page 1 of 3.

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.
(6)

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.
(6)

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.
(7)

Hardik said:   6 years ago
V = IR.
V directly proportional to I.
So, Apply proportional rules.
(1)

Ladli said:   7 years ago
Here, the load is constant(R).

So,v/i=R=v'/i'.
18/40mA=v'/20mA.
Hence, v'=9volt.

Swetha said:   7 years ago
Vg is directly proportional to I.
Ohm's law applies.
V1/V2=I1/I2 substitute this formula....

Yaswanth pagadala said:   8 years ago
Thanks for all.

Srk said:   8 years ago
v1=18 i1=40ma,i2=20ma,v2=?

R=18/40ma = 450.
V2 = i2 * r = 20m * 450 = 9a.

Jagan said:   9 years ago
Thanks for all the answer and explanation.
(1)

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.


Post your comments here:

Your comments will be displayed after verification.