# capa12 - Solution Derivations for Capa #12 1) An ideal...

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Unformatted text preview: Solution Derivations for Capa #12 1) An ideal step-down transformer has a primary coil of 300 turns and a secondary coil of 25 turns. It is plugged into an outlet with 120 V (AC) from which it draws a current of . 79 A . What is the voltage in the secondary coil? N p = Given N s = Given V p = Given (rms), subscript meaning primary I p = Given (rms) The given values for voltage and current are the rms values because you know it is plugged into an electrical outlet. From the ratio N s N p = V s V p we can find the voltage in the secondary coil (which is also an rms value). V s = V p N s N p 2) Calculate the rms current in the secondary coil. To find the rms current in the secondary coil, use the ratio I p I s = N s N p = V s V p or I s = I p N p N s Since the primary current I p was given as an rms quantity, the secondary current, I s , is also an rms value. 3) Assuming that the transformer secondary is driving a resistive load, calculate the average power dissipated in the resistor. Average power is calculated from the equation P = I rms V rms Where the rms current and rms voltage were calculated above. 4) The figure below shows the time variation of the current through an electrical heater when it is plugged into a 120 V , 60 Hz outlet. 1 V rms = Given f = Given The voltage is an rms quantity since you are told the device is plugged into an outlet. The frequency confirms that it is a household outlet. The relationship to rms voltage and peak voltage is V peak = 2 V rms 5) A power supply that provides a voltage of 49 . 0sin(388 t ) V is connected across a resistor R = 200 . Calculate the average power dissipated in the resistor. V ( t ) = Given R = Given The power dissipated through a resistor is P = I 2 rms R = V 2 rms R You are given an equation for the voltage. The peak voltage is simply the am- plitude of the sinusoidal function. The rms value is then calculated by V rms = V peak 2 So, P = V 2 rms R = 1 R V peak 2 ! 2 = V 2 peak 2 R 6) A source of electromagnetic waves radiates power uniformly in all directions at a single frequency. At a distance of 6 . 50 km from the source, a detector measures the intensity of the wave to be 20 . W/m 2 . What is the peak value of the magnetic field at the detector....
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## This note was uploaded on 04/03/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.

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capa12 - Solution Derivations for Capa #12 1) An ideal...

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