No it will not work.
Yes, a 3A 250V fuse can replace a 3A 125V fuse, as the voltage rating of the replacement fuse is higher than that of the original. The critical factor is that the current rating (3A) remains the same, ensuring it will blow under the same overload conditions. However, it's essential to ensure that the device or circuit is designed to operate safely within these ratings. Always follow manufacturer guidelines for fuse replacement to maintain safety and functionality.
Resistance = V/I Dimensional formula for V ML2T -3A -1 Dimensional formula for I A Dimensional formula for R= ML2T -3A -1 / A = ML2T -3A -2
Resistance = V/I Dimensional formula for V ML2T -3A -1 Dimensional formula for I A Dimensional formula for R= ML2T -3A -1 / A = ML2T -3A -2
V= (Area of base)(height) divide the whole thing by 3Both of these formulas work: V=1/3A*H or V=1/2L*H♥See related links for more details and illustrations.
According to ohm's law V = IR. So here I=V/R =12/4 I=3A
V = irv = (0.5)(250)v = 125
Where_is_locate_spark_plug_ford_2001_van_E-250
V=ir v= (250*10-6) a * (40*103) ohms v= ( .00025)a * 40000 ohms v= 10
For a 60W lamp, a normal fuse rating would be around 3A. This is calculated using the formula P = V x I, where P is power, V is voltage, and I is current. So for a 60W lamp on a 120V circuit, you would use a 3A fuse to safely handle the power load.
5xcube root of 2 or 5x(2^(1/3) which is about 6.29961 Or The volume of a cube, V, is: V = a^3 (side^3), we have: 250 = a^3 or a^3 = 250 a =(250)^(1/3) (you can do this in your calculator) a ≈ 6.29961
V=IR so 3a x 15 ohms gives 45v
Using Ohm's Law. It is 2.5 volts divided by 5.0 ohms which is 0.5 Amperes.