what is the resistance in a light bulb with a voltage of 25v and a current of 0.2a

R = 125Ω

Explanation:

Use the equation for voltage:

V = IR

V is voltage, in volts V

I is current, in amperes A

R is resistance, in ohms Ω

Substitute V for 25V and I for 0.2A

V = IR

(25V) = (0.2A)R  Isolate R for resistance. Divide both sides by 0.2

R = 125Ω

The resistance for the lightbulb is 125 Ω.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.