Question

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

Answer 1

Answer:

(a) 80.2 volt

(b) 113.40 volt

(c) 11.34 A

(D) 643.204 volt

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

$v_m=\sqrt{2}v_{rms}=1.414\times 80.2=113.40volt$

(c) Maximum current is given by

$i_m=\frac{v_m}{R}=\frac{113.40}{10}=11.34A$

(d) Average power delivered to resistor $P=VI=80.2\times 8.02=643.204watt$