An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
Determine the PIV rating as the peak secondary voltage: P I V = 12 V .
Calculate the DC current through the load: I d c = 370 π 12 ≈ 10.3 m A .
Compute the DC power supplied to the load: P d c = ( 370 π 12 ) 2 × 370 ≈ 39.4 mW .
Find the conversion efficiency: η = π 2 4 ≈ 40.5% . 12 V , 10.3 m A , 39.4 mW , 40.5%
Explanation
PIV Rating The peak inverse voltage (PIV) is the maximum reverse voltage that a diode can withstand without breaking down. In a half-wave rectifier, the PIV is equal to the peak secondary voltage, V m . Given v s = 12 sin ( ω t ) volts, the peak secondary voltage is 12V. Therefore, the PIV rating of the diodes is 12V.
DC Current The DC current through the load, I d c , is given by I d c = π I m , where I m is the peak current. The peak current is I m = R L V m = 370 12 . Therefore, I d c = 370 π 12 ≈ 0.0103 A = 10.3 m A .
DC Power The DC power supplied to the load, P d c , is given by P d c = I d c 2 R L = ( 370 π 12 ) 2 × 370 = 370 π 2 1 2 2 ≈ 0.0394 W = 39.4 mW .
Conversion Efficiency The conversion efficiency, η , is given by η = P a c P d c , where P a c is the AC power supplied to the rectifier. The AC power is P a c = 2 R L V m 2 = 2 × 370 1 2 2 . Therefore, η = 4 × 370 1 2 2 370 π 2 1 2 2 = π 2 4 ≈ 0.4053 . Converting to percentage, the conversion efficiency is approximately 40.53% .
Final Answer
PIV rating of the diodes: 12 V
DC current through the load: 10.3 mA
DC power supplied to the load: 39.4 mW
Conversion efficiency of the current: 40.5%
Examples
Half-wave rectifiers are commonly used in low-power applications such as battery chargers and simple DC power supplies. Understanding the PIV rating ensures the diode can withstand the reverse voltage, preventing breakdown. Calculating DC current and power helps determine the suitability of the rectifier for the load. Conversion efficiency is crucial for assessing the energy efficiency of the power supply, guiding design improvements for minimal energy waste.
A device delivering a current of 15.0 A for 30 seconds will have approximately 2.81 × 10^21 electrons flowing through it. This is calculated by finding the total charge using the formula Q = I × t and then converting that charge into the number of electrons based on the charge of a single electron. Therefore, the number of electrons is calculated to be around 2.81 × 10^21.
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