What is the power dissipated by an 8.2kΩ resistor if a current of 0.005 amps is passing through the resistor?

Answer 1

Using Ohm's law, E=IR, where E = the potential difference in volts, I the current in amperes, and R is the resistance in ohms.

Another formula gives the Power, P, =EI, where P is the power in watts, and E and I as given above.

Substituting, E=IR in the above equation, we get:

P=(IR)I or
P=I^2R.

Now substitute the given values :

P=(0.005)^2(8.2*1000)
= 0.000025*8.2*1000
=0.205 watt

Answer 2

Use P = I^2 R

Answer 3

P = I^2 * R

Answer 4

A fuse: A designed failure element in a circuit to negate destroying the full circuit to an over-load/short-circuit etc. Volts (V) = contemporary (I) x Restistance (R) Rearranging I = V / R ------------ (a million) Now R = Rho x length (L) / section (a) ---- (2) Substituting 2) right into a million) we get I = V x a / Rho x L --------------------(3) shall we make V = a million; Rho = a million and L = a million (all consistent) subsequently I = a As section “a” will boost “I” will boost too; as “a” decreases “I” decreases. by using the above a relentless fabric variety ought to be decreased in diameter over a length to create a failure element. in spite of if fuses may be made up of a diverse fabric then the circuit yet nevertheless rated to fail at a undeniable ampere cut back – take 13 amp fused plugs interior the united kingdom; based upon the ampage rigidity in circuits cables can replace into heat yet that degree relies upon on the layout traits. Fuses see problems in the previous those problems create destruction someplace else frequently.

Answer 5

The power dissipated is P=I^2 * R, where I is the current 0.005 amps, and R is the resistance 8.2k ohms.

Answer 6

P = I²R so .005²*8200 = .205 W ☺


Doug

Answer 7

p=i^2r=.005^2*8200=.205w

Answer 8

p = i^2R = 0.005^2*8.2 =0.205mW