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.

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)

Roopesh said:   2 years ago
Apply Ohm's law;
V1/V2 = I1/I2.
24/V2 = 120/160,
V2 = 24/0.75.
V2 = 32volts.
(2)

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)

AtirRH said:   1 decade ago
I = 120 mA (1 mA = 1/1000 A).
I = 120*1/1000 A or 12/100 A.

V = 24 v.
R = ?

According to ohm's law (V = IR).

R = V/I.
R = 24/12/100.
R = 200.

Again,

I = 160*1/1000 A or 16/100.
R = 200.
V = ?

According to ohm's law (V = IR).

V = 16/100*200.
V = 32 v.

AtirRH said:   1 decade ago
I = 120 ma to 160 ma.

How many times it increase?

I = 160/120.
I = 4/3 times.
V = 24*4/3.
V = 32 v.

DM Matsaba said:   10 years ago
Thank you all for giving different methods to get the answer.

M.srav said:   9 years ago
Thank you all for giving me an idea.

Sachin said:   9 years ago
Thank you @Ashutosh Sharma.

Sunil said:   9 years ago
120 : 160 : 24 : x.
x = 160 * 24/120,
x = 32.


Post your comments here:

Your comments will be displayed after verification.