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kari74 [83]
2 years ago
15

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

Physics
1 answer:
svp [43]2 years ago
8 0

(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.

<h3>Maximum potential difference</h3>

Vrms = 0.7071V₀

where;

  • V₀ is peak voltage

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

<h3> rms current through the resistor </h3>

I(rms) = V(rms)/R

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

I(rms) = 0.16 A

<h3>maximum current through the resistor </h3>

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

<h3> Average power dissipated by the resistor</h3>

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

#SPJ1

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mariarad [96]

Answer:

Gypsum

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Hope this helped:)

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3 years ago
If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out
Maru [420]

Answer:

power=work done÷time taken

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10÷10=1

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1 year ago
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6 0
2 years ago
Read 2 more answers
The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not
Nuetrik [128]

Answer:

P(bat) = V²r/(R+r)²

Explanation:

Let the resistance of the coil be R

Internal resistance of the battery be r

Emf of the battery = V

Power dissipated in the internal resistance of the battery is normally given as P = I²r

where I is the current flowing in the circuit.

From Ohm's law,

V = I R(eq)

R(eq) = (R + r)

I = V/(R+r)

P = I²r

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Hope this Helps!!!

6 0
2 years ago
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