Question

2. A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.

Answer 1

(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

Maximum potential difference

Vrms = 0.7071V₀

where;

• V₀ is peak voltage

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

rms current through the resistor

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

maximum current through the resistor

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

Average power dissipated by the resistor

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

Learn more about maximum current here: brainly.com/question/14562756

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