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.
Roderick acuesta said:
9 years ago
The problem is not clear because not tell if the load is variable.
Sunil Pant said:
1 decade ago
40mA -> 18V
Then 20mA -> (?)
20*18/40 = 9V
Then 20mA -> (?)
20*18/40 = 9V
Jagan said:
9 years ago
Thanks for all the answer and explanation.
(1)
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
Nikul Prajapati said:
1 decade ago
40mA - 18V
then 20mA - (?)
20*18/40 = 9V
then 20mA - (?)
20*18/40 = 9V
Vinodroyal said:
9 years ago
If current increase voltage decreases.
S k s said:
9 years ago
Voltage change 18 - 9 = 9 volt.
Julea said:
1 decade ago
How did they get 9v?
Yaswanth pagadala said:
8 years ago
Thanks for all.
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