Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 14)
14.
If you wish to increase the amount of current in a resistor from 120 mA to 160 mA by changing the 24 V source, what should the new voltage setting be?
Discussion:
33 comments Page 2 of 4.
Raghu said:
1 decade ago
For 1v 5ma changed so like that 40 ma changes means 8 v changes.
So to add that 24+8=32v for 160A
So to add that 24+8=32v for 160A
Nikul Prajapati said:
1 decade ago
With help of maths therasuit
120mA - 24V
Then 160mA - (?) (How Much)
Means 160*24/120 = 32V
120mA - 24V
Then 160mA - (?) (How Much)
Means 160*24/120 = 32V
Nagnaga said:
1 decade ago
Hi santosh its obviously ohms law, since we are doing problems under ohms law subtitle.
Santhosh said:
1 decade ago
Hi nikul which formula you are using to solve this, may I know the name of the formula.
Roopesh said:
2 years ago
Apply Ohm's law;
V1/V2 = I1/I2.
24/V2 = 120/160,
V2 = 24/0.75.
V2 = 32volts.
V1/V2 = I1/I2.
24/V2 = 120/160,
V2 = 24/0.75.
V2 = 32volts.
(2)
Vinay babu said:
1 decade ago
V is directly proportional to I (i.e) go with @Ashutosh is very smart.
Sridhar said:
9 years ago
Just apply the formula V1/I1 = V2/I2 then we will get the answer.
DM Matsaba said:
10 years ago
Thank you all for giving different methods to get the answer.
Abdi sadick said:
9 years ago
Thanks for all your calculation. I get it now.
Ashutosh Sharma said:
1 decade ago
V1/I1 = V2/I2.
24/120 = V2/160.
V2 = 32Volt.
24/120 = V2/160.
V2 = 32Volt.
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