Problem 31.3: Power and resistive loads Please wait for the animation to complet

Question

Problem 31.3: Power and resistive loads Please wait for the animation to completely load. Assume an ideal power supply. The graph shows the voltage across the source (red) and the current through the circuit (blue) as functions of time (voltage is given in volts, current is given in milliamperes [10^-3 amperes], and time is given in seconds). Restart. For each circuit, What is the rms voltage? What is the rms current? What is the value of the unknown resistor? What is the average power dissipated?

Explanation / Answer

Here ,

peak voltage is given as

Vpeak = 5 V

Now, rms voltage = Vpeak/sqrt(2)

rms voltage = 5/sqrt(2)

rms voltage = 3.54 V

the rms voltage is 3.54 V

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for rms current

peak current = 3 mA

rms curernt = 3 *10^-3/sqrt(2)

rms curernt = 2.121 mA

the rms curernt is 2.121 mA

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using ohm's law

I = V/R

R = 3.54/(2.121 *10^-3)

R = 1669 Ohm

the resistance is 1669 Ohm

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Power dissipated = 3.54 * 2.121 *10^-3

Power dissipated = 7.508 *10^-3 W

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