The device will work, but the 3A fuse will blow quicker than the 5A would have.
7a minus 2, if "3a-2" means 3a minus 2. 3a plus 1 and 1-3rd in parenthesis times 3a, if "3a-2" means 3a squared. a plus 3a squared plus 3a = 1-3rd times 3a plus 3a times 3a plus 1 times 3a = 3a plus 1 and 1-3rd in parenthesis times 3a
p=b+3a+c p-3a-c=b+3a-3a+c-c p-3a-c=b b=p-3a-c
Simplifying a = 3a + 12 Reorder the terms: a = 12 + 3a Solving a = 12 + 3a Solving for variable 'a'. Move all terms containing a to the left, all other terms to the right. Add '-3a' to each side of the equation. a + -3a = 12 + 3a + -3a Combine like terms: a + -3a = -2a -2a = 12 + 3a + -3a Combine like terms: 3a + -3a = 0 -2a = 12 + 0 -2a = 12 Divide each side by '-2'. a = -6 Simplifying a = -6
(3 x 3) - 6 = 3 3a-6=3 3a=3+6 3a=9 a=9/3 a=3
No
Yes, fuses of a higher voltage can be used safely. What is not recommended is to use a lower voltage fuse on higher voltages.
No, it is not recommended to use a 3.15A fuse to replace a 3A fuse. The 3.15A fuse has a higher current rating and may not provide adequate protection for the circuit. It is best to replace a fuse with the same current rating to ensure proper circuit protection.
The device will work, but the 3A fuse will blow quicker than the 5A would have.
A: That would be OK ONLY if the switch is to carry 3 amps maximum meaning 120v AC or approximately 40 watts
The fuse will blow as soon as you turn it on. Use the correct fuse and nothing else.
If you do that the likelihood is that you will blow the 3A fuse quickly. There is a reason why the current fuse is what it is, because it is expecting currents around 80% of 13 A or around 10 A.
You could replace it with a 3A fuse. You should never replace a fuse with one which is rated higher.
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.
To calculate the minimum fuse rating needed for a 36W bulb on a 12V circuit, divide the wattage by the voltage (36W / 12V = 3A). Therefore, a minimum 3A fuse would be sufficient for a 36W bulb on a 12V circuit.
3a^2 + 3a^2 = 6a^2 3a^2 - 3a^2 = 0 3a^2 x 3a^2 = 9a^4 3a^2 divided by 3a^2 = 1
7a minus 2, if "3a-2" means 3a minus 2. 3a plus 1 and 1-3rd in parenthesis times 3a, if "3a-2" means 3a squared. a plus 3a squared plus 3a = 1-3rd times 3a plus 3a times 3a plus 1 times 3a = 3a plus 1 and 1-3rd in parenthesis times 3a