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 2 of 3.

Estrellanes said:   9 years ago
The half of 40 is 20, and the half of 18 is 9. So the answer is A) 9v.

S k s said:   9 years ago
Voltage change 18 - 9 = 9 volt.

Roderick acuesta said:   9 years ago
The problem is not clear because not tell if the load is variable.

Vinodroyal said:   9 years ago
If current increase voltage decreases.

Krishna said:   9 years ago
Or you can say V is directly proportional to I. So when the current drops so as the voltage.

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.

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.

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.

Julea said:   1 decade ago
How did they get 9v?

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.


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