This is an insidious question. Quite frankly, I would not have
expected to see it here on Brainly. But I'm ready to play the
cards that you have dealt me.
None of the choices offered is a correct solution.
If the output of the AC generator is nice and sinusoidal, and
its maximum (peak) emf is 150 volts, then its RMS emf is
(1/2) (150) (√2) = 106.07 volts.
The resistor's dissipation is
Power = (current) x (voltage) .
If the resistor is dissipating its full rated 35W, then
35W = (current) x (106.07 V)
Divide each side by 106.07 V:
RMS Current = (35W) / (106.07 V) = 0.33 Ampere .
_________________________________________
Looking over the choices offered . . .
The largest choice ... 3.1 A ... is the current in a resistor
that is dissipating 35W if the voltage is
(35W / 3.1A) = 11.29 volts .
The smallest choice ... 1.2 A ... is the current in a resistor
that is dissipating 35W if the voltage is
(35W / 1.2A) = 29.17 volts .
Whatever you meant the so-called "150 V" of the generator
to represent ... whether the RMS sinusoidal, peak sinusoidal,
peak square-wave, RMS square-wave, DC, average, etc. ...
none of the choices for current, in combination with any of these
generators, would dissipate 35W.
Having the ability to deal with people in a way that does not offend them
Answer:
Explanation:
The linear charge density id defined as the charge per unit length.
Let q be the charge and L be the length of the wire.
So, the linear charge density is
λ = q / L
If the length is 12 times, then charge should also be 12 times to keep the linear charge density same.
