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 3 of 4.
Sachin said:
9 years ago
Thank you @Ashutosh Sharma.
Sunil said:
9 years ago
120 : 160 : 24 : x.
x = 160 * 24/120,
x = 32.
x = 160 * 24/120,
x = 32.
Sridhar said:
9 years ago
Just apply the formula V1/I1 = V2/I2 then we will get the answer.
Suri said:
9 years ago
Logically you are right @Raghu.
Abdi sadick said:
9 years ago
Thanks for all your calculation. I get it now.
Shahid ansari said:
8 years ago
Good trick, thank you @Ashutosh Sharma.
Krunal soni said:
8 years ago
Nice trick, Thaks @Ashutosh.
Chuki said:
7 years ago
Change 120mA to 0.12A.
Then solve for R. R = (24/0.12) = 200 ohms.
So, it would be R = 200 ohms, I = 0.16A. V = IR = (0.16)(200) = 32 v.
Then solve for R. R = (24/0.12) = 200 ohms.
So, it would be R = 200 ohms, I = 0.16A. V = IR = (0.16)(200) = 32 v.
(1)
Swetha said:
7 years ago
Ohm's law applies.
V1/V2=I1/I2.
V1/V2=I1/I2.
(1)
Jake said:
7 years ago
Thanks @Mani.
The resistance is constant.
The resistance is constant.
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