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.
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.
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)
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 "...
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.
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:
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.
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.
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.
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.
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.
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.
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)
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.
24/120 =V2 /160,
8/1 = V2/4,
V2= 32 Volt.
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.
TAKE RATIO V1/I1 = V2/I2.
V2 = (V1*I2)/I1.
V2 = (24*160)/120.
V2 = (3840)/120.
V2 = 32V is answer.
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.
How many times it increase?
I = 160/120.
I = 4/3 times.
V = 24*4/3.
V = 32 v.
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