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?
8 V
320 V
3.2 V
32 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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.
(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.
(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.

Swetha said:   7 years ago
Ohm's law applies.

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.
(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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