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 4 of 4.
Vaidya Preet said:
6 years ago
Thank You for the explanation of the answer.
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)
Roopesh said:
2 years ago
Apply Ohm's law;
V1/V2 = I1/I2.
24/V2 = 120/160,
V2 = 24/0.75.
V2 = 32volts.
V1/V2 = I1/I2.
24/V2 = 120/160,
V2 = 24/0.75.
V2 = 32volts.
(2)
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