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.
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.
Jake said:
7 years ago
Thanks @Mani.
The resistance is constant.
The resistance is constant.
Vaidya Preet said:
6 years ago
Thank You for the explanation of the answer.
Ajith128 said:
1 decade ago
No need to go for calculation 4 like these maths. We know only option to increase current in a give circuit increase voltage. So to increase current voltage must be increased. All other values are less the 24 and only 32 and 320 are higher.
AtirRH said:
1 decade ago
I = 120mA (1mA = 1000 A).
I = 120*1000 A or 12*10000 A.
V = 24v.
R = ?
According to ohm's law (V = IR).
R = V/I.
R = 24/12*10000.
Again,
I = 160*1000A or 16*10000.
R = 24/12*10000.
V = ?
According to ohm's law (V = IR).
V = 16*10000*24/12*10000.
V = 32v.
I = 120*1000 A or 12*10000 A.
V = 24v.
R = ?
According to ohm's law (V = IR).
R = V/I.
R = 24/12*10000.
Again,
I = 160*1000A or 16*10000.
R = 24/12*10000.
V = ?
According to ohm's law (V = IR).
V = 16*10000*24/12*10000.
V = 32v.
Golam Mahmud said:
1 decade ago
v1/v2 = I1/I2.
So v2 = 32V.
So v2 = 32V.
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