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.
Ajith128 said:
1 decade ago
No need to go for calculation 4 like these maths. We know only option to increase current in a give circuit increase voltage. So to increase current voltage must be increased. All other values are less the 24 and only 32 and 320 are higher.
Raghu said:
1 decade ago
For 1v 5ma changed so like that 40 ma changes means 8 v changes.
So to add that 24+8=32v for 160A
So to add that 24+8=32v for 160A
MANI said:
1 decade ago
1ST INCREASING THE AMOUNT OF CURRENT FROM 120mA THEN CHANGING THE VOLTAGE FROM 24V.. K SO, R=V/I R=24/120 R=0.2K.O ... NOW WE R FIND RESISTOR VALUE... NOW WE WANT NEW VOLTAGE SETTING... SO,PUT THE RESISTOR VALUE IN OHMS LAW... ALREADY WE HAVE mA VALUE.. SO, V=I*R V=160*0.2 "V=32V "...
Pooms said:
1 decade ago
Thank you Mr.Mani...... I'm very clear....
Nikul Prajapati said:
1 decade ago
With help of maths therasuit
120mA - 24V
Then 160mA - (?) (How Much)
Means 160*24/120 = 32V
120mA - 24V
Then 160mA - (?) (How Much)
Means 160*24/120 = 32V
Sibananda chakra said:
1 decade ago
Thank you all of you. I am very clear.
Prashant said:
1 decade ago
Nice calculation of nikul.
Santhosh said:
1 decade ago
Hi nikul which formula you are using to solve this, may I know the name of the formula.
Nagnaga said:
1 decade ago
Hi santosh its obviously ohms law, since we are doing problems under ohms law subtitle.
Ashutosh Sharma said:
1 decade ago
V1/I1 = V2/I2.
24/120 = V2/160.
V2 = 32Volt.
24/120 = V2/160.
V2 = 32Volt.
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