1. Electricity: What’s the connection?
PSC 4011
Electricity:
What’s the connection?

2. Series Parallel Combined
PSC 4011: Power & Energy
Series Parallel Combined

3. PSC 4011: Power & Energy

4. PSC 4011: Power & Energy
Power: Is the quantity of energy produced or delivered per second.
Symbol: P
Unit: Watts (W)
𝑷=𝑽×𝑰 (which can be measured directly from the circuit by using an ammeter and voltmeter)

5. PSC 4011: Power & Energy
Power delivered = Power consumed ( E conservation) In any given circuit (series, parallel or combined) the power delivered by the power supply is equal to the sum of the power consumed by each of the energy consuming devices P = ε It = P1 + P2 … _Parallel: ε = V1 = V2…, It = I1+ I2 … _Series: ε = V1 + V2…, It = I1 = I2 …

6. PSC 4011: Power & Energy
Practice Calculate the power supplied by a 9V-battery that draws a current of 30 mA. What would be the power consumed by the circuit fed by said battery? P = VI P = (9V)(0.03A) P = 0.27 W
Power consumed by circuit is also 0.27 W
(P delivered = P consumed)

7. PSC 4011: Power & Energy
Joule effect: Heat generated when a current flows through a resistor since electrical energy is converted into heat. Formula: 𝑷=𝑰𝟐×𝑹 since V = I x R (Ohm’s Law) *The higher the electrical current (I) the higher the heat generated, and so the higher the power / energy loss since electrical energy is transformed into heat

8. PSC 4011: Power & Energy
Fuses or circuit breakers have been introduced due to the Joule effect.
Fuses or circuit breakers work as switches that melt (fuses) or open themselves (breakers) and stop current circulation, thus avoiding major disasters.

9. PSC 4011: Power & Energy
Thinner conductors, having higher values of resistance, have lower limits of permissible current for energy transportation.
Conversely, thicker wires are permitted to transport higher values of current intensity.

10. PSC 4011: Power & Energy
Practice Calculate the power lost due to JE in a 2km aluminum conductor with R = 5.6 Ω/km through which a 5A current passes. Would you use a thicker or a thinner conductor in order to decrease the loss? P = RI2 P = (5.6 Ω/km * 2 km)(5A) 2 P = 280 W
I would rather use a thicker conductor, in order to decrease R and so the power loss due to JE

11. PSC 4011: Power & Energy
Another way to minimize losses due to JE is by increasing voltage for energy transportation (use of higher tension wires) while keeping the power constant (thus decreasing current intensity, and power loss due to Joule Effect)

12. PSC 4011: Power & Energy
Practice A power plant puts out 50 kV, but said voltage is increased to 200 kV for transportation. Calculate the reduction of power loss if power delivered by the plant is kept at 100 kW, and the conductors have a R = 5.6 Ω/km for a trajectory of 5 km. P1 = RI2 = R (P/V) 2 P1 = (5.6 Ω/km * 5 km)(100kW/50kV) 2 P1 = 112 W P2 = RI2 = R (P/V) 2 P2 = (5.6 Ω/km * 5 km)(100kW/200kV) 2 P2 = 7 W Reduction of power loss = 112 W – 7 W = 105 W

13. PSC 4011: Power & Energy
If resistance is lowered, but voltage is kept at same value, current intensity will increase and thus losses due to Joule effect will be greater

14. PSC 4011: Power & Energy
Practice Calculate how much more loss due to Joule effect will be produced if resistance of device is halved, while voltage is kept at same value. What would happen if resistance were reduced to a half of the original value, while voltage would still be constant? V = IR I = V/R I’ = V/R/2 I’ = 2 V/R I’ = 2 I
P = RI2 P’ = R’I’2 P’ = (R/2)(2I)2 P’ = 4(R/2)I2 P’ = 2 RI2 P’ = 2 P (if R is reduced to a half, but V is kept constant, P loss will double, since I will double) P” = 3 P (if R is reduced to a third, but V is kept constant, P loss will triple, since I will triple)

15. PSC 4011: Power & Energy
Energy: The amount of energy used by an electrical device is often the same as the work done by the device. Symbol: E Unit: Joule (J) Energy can be calculated using power and time: 𝑬=𝑷×𝒕

16. PSC 4011: Power & Energy
Energy delivered = Energy consumed ( E conservation) In any given circuit (series, parallel or combined) the energy delivered by the power supply is equal to the sum of the energy consumed by each of the energy consuming devices E = (ε It) t = E1 + E2 … _Parallel: ε = V1 = V2…, It = I1+ I2 … _Series: ε = V1 + V2…, It = I1 = I2 …

17. PSC 4011: Power & Energy
Practice Determine the amount of energy consumed by a 700W-microwave oven used for ten minutes. E = Pt E = (700 W)(600 s) E = J or 420 kJ

18. PSC 4011: Power & Energy
The kilowatt-hour is the amount of energy consumed by a device that uses 1 kW in one hour. E = Pt E = (1000 W)(3600 s) E = J or 3.6 MJ

19. PSC 4011: Power & Energy
Practice Determine the amount of energy consumed by a 700W-microwave oven used for ten minutes. Express it in kWh E = Pt E = (0.7 kW)(600 s) / (3600 s) E = 0.12 kWh

20. PSC 4011: Power & Energy
Energy efficiency: Percentage of useful energy out of the total energy generated by a system. 𝐸 𝑒𝑓𝑓 = 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐸 𝑢𝑠𝑒𝑓𝑢𝑙 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐸 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 ×100 * E useful = E generated – E lost (Joule Effect)

21. PSC 4011: Power & Energy
Practice Calculate the increase in energy efficiency for a plant that reduces energy losses (due to Joule effect) from 500 J to 250 J, if the total amount of energy generated in each case is the same (5000 J). Eeff = (Euseful / Egenerated) x 100 Eeff = (Egenerated - Elost / Egenerated) x 100 Eeff = (5000 J – 500 J / 5000 J) x 100 = 90 % Eeff = (5000 J – 250 J / 5000 J) x 100 = 95 % There is a 5 % increase in energy efficiency

22. PSC 4011: Power & Energy
Power efficiency: Percentage of useful power out of the total power generated by a system. 𝑃 𝑒𝑓𝑓 = 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑃 𝑢𝑠𝑒𝑓𝑢𝑙 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑃 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 ×100 * P useful = P generated – P lost (Joule Effect)

23. PSC 4011: Power & Energy
Practice Calculate the increase in power efficiency for a plant that reduces power losses (due to Joule effect) from 1112 W to 70 W, if the total amount of power generated in each case is the same ( J). Peff = (Puseful / Pgenerated) x 100 Peff = (Pgenerated - Plost / Pgenerated) x 100 Peff = ( W – 1112 W / W) x 100 = % Peff = ( W – 70 W / W) x 100 = % There is a 1.05 % increase in power efficiency