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 2 of 4.

Sridhar said:   9 years ago
Just apply the formula V1/I1 = V2/I2 then we will get the answer.

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

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

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

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

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.

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.

Golam Mahmud said:   1 decade ago
v1/v2 = I1/I2.

So v2 = 32V.

Jaswinder singh said:   1 decade ago
V = IR.
TAKE RATIO V1/I1 = V2/I2.
V2 = (V1*I2)/I1.
V2 = (24*160)/120.
V2 = (3840)/120.
V2 = 32V is answer.


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