Question

One cycle of the power dissipated by a resistor (R=800 Ω) is given by

P(t)=0.03125 W, 0≤t<12.0 sP(t)=0.03125 W, 0≤t<12.0 s
P(t)=0 W, 12.0≤t<30 sP(t)=0 W, 12.0≤t<30 s
This periodic signal repeats in both directions of time.
What is the amplitude of the PWM voltage signal applied across the 800-Ω resistor?

Answer 1

Answer:

Maximum voltage = 5 V

Explanation:

The maximum power from the resistor is evidently 0.03125W in the time periods given.

And Power dissipated from electric circuits is given by

P = IV

V = IR

P = I(IR) = I²R

Since I = (V/R), power can also be expressed as

P = IV = (V/R)V = V²/R

Since the resistance of the resistor isn't variable, then the variable here has to be the voltage, So, to calculate maximum voltage,

R = 800 ohms, P = 0.03125 W

0.03125 = V²/800

V² = 25

V = 5V

Answer 2

In this exercise we have to use the knowledge of voltage to calculate its final value, so we have to:

The value Maximum voltage = 5 V

Recalling some formulas used to calculate the voltage of a resistor we have that:

$P = IV\\V = IR\\P = I(IR) = I^2R\\I = (V/R)\\P = IV = (V/R)V = V^2/R$

They are putting the values ​​that are already known in the formulas is:

• R = 800 ohms
• P = 0.03125 W

We have that will be equal to:

$0.03125 = V^2/800\\V^2 = 25\\V = 5V$

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