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 1 of 4.
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)
Aakarsh Chandra said:
3 years ago
We'll first find the Resistance.
Let's find it
V=IR,
R=V/I.
R=24/120*10^-3,
R=24*1000/120,
R=24000/120,
R=200 (Ohm).
And now we get the value of resistance that is 200 ( Ohm ).
But we need 160 mA of current that's why now we'll find the value of Voltage and how much voltage is increasing.
Let's find,
V=IR.
V= 160*10^-3 * 200,
V= 160*200/1000,
V= 32000/1000,
V=32v.
THANK YOU.
Let's find it
V=IR,
R=V/I.
R=24/120*10^-3,
R=24*1000/120,
R=24000/120,
R=200 (Ohm).
And now we get the value of resistance that is 200 ( Ohm ).
But we need 160 mA of current that's why now we'll find the value of Voltage and how much voltage is increasing.
Let's find,
V=IR.
V= 160*10^-3 * 200,
V= 160*200/1000,
V= 32000/1000,
V=32v.
THANK YOU.
(6)
Vaidya Preet said:
6 years ago
Thank You for the explanation of the answer.
Jake said:
7 years ago
Thanks @Mani.
The resistance is constant.
The resistance is constant.
Swetha said:
7 years ago
Ohm's law applies.
V1/V2=I1/I2.
V1/V2=I1/I2.
(1)
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)
Krunal soni said:
8 years ago
Nice trick, Thaks @Ashutosh.
Shahid ansari said:
8 years ago
Good trick, thank you @Ashutosh Sharma.
Abdi sadick said:
9 years ago
Thanks for all your calculation. I get it now.
Suri said:
9 years ago
Logically you are right @Raghu.
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