The alternating current through the 15 Ω heating element of a toaster produces power at 980 W. Calculate (a) the rms current through the element, (b) peak current, (c) the rms voltage across the element, and (d) the peak voltage.
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The alternating current through the 15 Ω heating element of a toaster produces power at 980 W. Calculate (a) the rms current through the element, (b) peak current, (c) the rms voltage across the element, and (d) the peak voltage.
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- The peak value of an alternating current in a 1800 −W device is 5.8 A. What is the rms voltage across it? Express your answer using two significant figures.In operation an electric toaster has an rms current of 3.95 A when plugged into a wall socket with rms voltage 1.20 x 102 V. Find the toaster’s (a) average power, (b) maximum instantaneous power, and (c) resistance.The peak value of an alternating current in a 1500 W device is 5.4 A. What is the rms voltage across it?
- (a) Old incandescent lightbulbs have a purely resistive metal filament, through which flows an alternating current. If an incandescent bulb uses an average power of 60.0 W when connected to a 60.0 Hz power source having a maximum voltage of 170 V, what is its resistance (in Ω)? Ω (b) Another incandescent bulb, with an average power usage of 75.0 W, is connected to the same AC power source, with the same frequency and maximum voltage. What is its resistance (in Ω)? ΩA microphone has a resistance of 7.8 Ω. When the microphone is connected to a voltage source, the average power it dissipates is 89 W. (a) What is the rms current? in A(b) What is the peak voltage?A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V. At what rms voltage will the resistor dissipate 10.0 W?
- The peak‑to‑peak current passing through a 130-ohm resistor is 28.0 A. Find the maximum voltage V0 across the resistor. Find the rms current i(rms) through the resistor.An A.C. circuit consists of an alternating voltage source and a resistor only. In this circuit ______. the voltage is in phase with the currentthe voltage leads the current by 90° the voltage lags the current by 180° the voltage lags the current by 90° the voltage leads the current by 180°Example: (a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak +V, power consumption rate of a 60.0-W AC light bulb? For (a): -Vo Solving the equation Vrms= for the peak voltage Vo and substituting the known value for Vms gives: Vo=2-vZVrms=1.414(120 V)= 170 V. This means that the AC voltage swings from 170 V to -170 V and back 60 times every second. An equivalent DC voltage is a constant 120 V. For (b): Peak power is peak current times peak voltage. Thus, Figure 2. The potential difference V between the terminals of an AC voltage So the power swings from zero to 120 W on hundred twenty times per second (twice each cycle), and the power averages 60 W. fluctuates shown. The Po= loVo = 2(; loVo) = 2Pave. source as mathematical expression for V is given by V = Vosin2nft Po = 2(60.0 W) = 120 W. Average power Activity: Given a transmission line having a line resistance of 5.000 sove for the following. Use the rubrics as your guide in presenting your solution…
- A 60-Hz AC generator develops a peak voltage of 170-V. The generator is connected to a lightbulb whose resistance when operating is 240Ω. Calculate an expression for the variation of current through the bulb with time. Use the formula: i=Vo/R sin(2πft)(a) Old incandescent lightbulbs have a purely resistive metal filament, through which flows an alternating current. If an incandescent bulb uses an average power of 40.0 W when connected to a 60.0 Hz power source having a maximum voltage of 170 V, what is its resistance (in 2)? 723 Ω (b) Another incandescent bulb, with an average power usage of 120 W, is connected to the same AC power source, with the same frequency and maximum voltage. What is its resistance (in 2)? 241Suppose you accidentally touched wires connected to the 120-V output from the wall socket. (a) calculate the rms current through your body if the resistance between electrodes is 10,000 Ω. (b)Calculate the peak current.