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A voltmeter and an ammeter are used to respectively monitor the voltage across and the current through a resistor connected to an AC source. If the resistor has a value of 10.0 Ω and the ammeter reads 8.02 A, determine the following.

(a) the rms voltage (in V) across the resistor(b) the peak voltage (in V) of the source(c) the maximum current (in A) in the resistor(d) the average power (in W) delivered to the resistor

1 Answer

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Answer:

(a) 80.2 volt

(b) 113.40 volt

(c) 11.34 A

(D) 643.204 volt

Step-by-step explanation:

We have given value of resistance R = 10 ohm

Current measured by the ammeter i = 8.02 A

(a) Rms voltage is given by

Voltage = resistance × current = 10×8.02 = 80.2 volt

(b) Peak voltage is equal to


v_m=√(2)v_(rms)=1.414* 80.2=113.40volt

(c) Maximum current is given by


i_m=(v_m)/(R)=(113.40)/(10)=11.34A

(d) Average power delivered to resistor
P=VI=80.2* 8.02=643.204watt

User Jason Watkins
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