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 3 of 4.

Ananda kumar said:   1 decade ago
They given 2 I ratings and V1 rating and asked to find the v2 value.
R = V/I.

V1/I1 = 24/120 = 8=R2.....For V1--------1.
V2/I2 = v2/160 = V2/4 = R2..For V2--------2.

Equate 2 Epns,
8 = V2/4
V2 = 32 volts.

Raju giri said:   1 decade ago
Here is two currents rating then we apply ohms for two voltage rating like V1/I1 = V2/I2 So,

24/120 =V2 /160,
8/1 = V2/4,
V2= 32 Volt.

Vinay babu said:   1 decade ago
V is directly proportional to I (i.e) go with @Ashutosh is very smart.

Ashutosh Sharma said:   1 decade ago
V1/I1 = V2/I2.

24/120 = V2/160.

V2 = 32Volt.

Nagnaga said:   1 decade ago
Hi santosh its obviously ohms law, since we are doing problems under ohms law subtitle.

Santhosh said:   1 decade ago
Hi nikul which formula you are using to solve this, may I know the name of the formula.

Prashant said:   1 decade ago
Nice calculation of nikul.

Sibananda chakra said:   1 decade ago
Thank you all of you. I am very clear.

Nikul Prajapati said:   1 decade ago
With help of maths therasuit

120mA - 24V

Then 160mA - (?) (How Much)

Means 160*24/120 = 32V

Pooms said:   1 decade ago
Thank you Mr.Mani...... I'm very clear....


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