# Electrical Engineering - Ohm's Law - Discussion

### Discussion :: 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?

 [A]. 8 V [B]. 320 V [C]. 3.2 V [D]. 32 V

Explanation:

No answer description available for this question.

 Ajith128 said: (Jan 5, 2011) 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: (Mar 3, 2011) For 1v 5ma changed so like that 40 ma changes means 8 v changes. So to add that 24+8=32v for 160A

 Mani said: (Mar 13, 2011) 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: (May 19, 2011) Thank you Mr.Mani...... I'm very clear....

 Nikul Prajapati said: (May 25, 2011) With help of maths therasuit 120mA - 24V Then 160mA - (?) (How Much) Means 160*24/120 = 32V

 Sibananda Chakra said: (Jul 6, 2011) Thank you all of you. I am very clear.

 Prashant said: (Jul 28, 2011) Nice calculation of nikul.

 Santhosh said: (Aug 30, 2011) Hi nikul which formula you are using to solve this, may I know the name of the formula.

 Nagnaga said: (Sep 16, 2011) Hi santosh its obviously ohms law, since we are doing problems under ohms law subtitle.

 Ashutosh Sharma said: (Jun 18, 2012) V1/I1 = V2/I2. 24/120 = V2/160. V2 = 32Volt.

 Vinay Babu said: (Oct 11, 2012) V is directly proportional to I (i.e) go with @Ashutosh is very smart.

 Raju Giri said: (Feb 24, 2014) 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.

 Ananda Kumar said: (Mar 6, 2014) 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.

 Jaswinder Singh said: (Aug 14, 2014) V = IR. TAKE RATIO V1/I1 = V2/I2. V2 = (V1*I2)/I1. V2 = (24*160)/120. V2 = (3840)/120. V2 = 32V is answer.

 Golam Mahmud said: (May 9, 2015) v1/v2 = I1/I2. So v2 = 32V.

 Atirrh said: (Jul 12, 2015) 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.

 Atirrh said: (Jul 12, 2015) 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: (Jul 19, 2015) 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.

 Dm Matsaba said: (May 18, 2016) Thank you all for giving different methods to get the answer.

 M.Srav said: (Jul 7, 2016) Thank you all for giving me an idea.

 Sachin said: (Aug 19, 2016) Thank you @Ashutosh Sharma.

 Sunil said: (Sep 12, 2016) 120 : 160 : 24 : x. x = 160 * 24/120, x = 32.

 Sridhar said: (Sep 21, 2016) Just apply the formula V1/I1 = V2/I2 then we will get the answer.

 Suri said: (Nov 6, 2016) Logically you are right @Raghu.

 Abdi Sadick said: (Apr 2, 2017) Thanks for all your calculation. I get it now.

 Shahid Ansari said: (Jun 14, 2017) Good trick, thank you @Ashutosh Sharma.

 Krunal Soni said: (Jun 17, 2017) Nice trick, Thaks @Ashutosh.

 Chuki said: (Jul 1, 2018) 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.

 Swetha said: (Aug 10, 2018) Ohm's law applies. V1/V2=I1/I2.

 Jake said: (Feb 13, 2019) Thanks @Mani. The resistance is constant.

 Vaidya Preet said: (Jul 30, 2019) Thank You for the explanation of the answer.