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Electronics maths

 

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1. A potential difference of 10 V exists across an entity. In 10^-6 seconds10^-9 Coulomb of charge flows into the entity and moves from the surfacewith a higher potential to the surface with the lower potential.a) Determine the energy dissipated in the entity.b) Determine the average power dissipated in the... 顯示更多 1. A potential difference of 10 V exists across an entity. In 10^-6 seconds 10^-9 Coulomb of charge flows into the entity and moves from the surface with a higher potential to the surface with the lower potential. a) Determine the energy dissipated in the entity. b) Determine the average power dissipated in the entity. 2. A copper wire has a conductivity of 10^6 AV^–1m^–1, a radius of 1mm and a length of 1 meter. a) Determine the resistance of the wire. b) If a voltage of 0.01 volts is placed across the wire then determine the current that will flow. 3. A voltage of 15 volts is placed across a resistance of 50Ω. Calculate: a) The current flowing through the resistance. b) The power dissipated in the resistance. 4. A current of 3ma flows through a resistance of 1kΩ. Calculate: a) The voltage drop across the resistance. b) The power dissipated in the resistance.

最佳解答:

1.a. Energy dissipated, E = QV = (10-9)(10) = 10-8 J b. Average power dissipated = E / t = 10-8 / 10-6 = 10-2 W 2.a. Resistance, R = l / CA where l is the length of the wire, C is the conductivity of the wire A is the area of the wire So, R = (1) / (106)π(0.001)2 R = 0.318 Ω b. Current, I = V / R = 0.01 / 0.318 = 0.0314 A 3.a. Current, I = V / R = 15 / 50 = 0.3 A b. Power dissipated, P = IV = (0.3)(15) = 4.5 W 4.a. Voltage drop, V = IR = (0.003)(1000) = 3 V b. Power dissipated, P = I2R = (0.003)2(1000) = 0.009 W = 9 X 10-3 W

其他解答:0D7DAC4E6DF60DFA

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