Question

(C) Using a circuit analysis technique of your choosing, calculate the power balance of the system at midday. Show all working. Comment on the effectiveness (calculate percentages) of power transfer from the solar panel to the LED lights, and how much is dissipated in the wiring. If LED brightness is proportional to current, express the relative brightness of the LED lights if LED 3 is used as a reference (100%),

a=560

b=0.92

c=28

d=2.8

e=2.4

f=6.6

g=31.5

(d) Use circuit simulation software to verify your calculations in (c). Capture screenshots of the software showing displayed values of voltages and currents that confirm your results.

Sarah the sommelier wants to use a skylight to brighten the dim wine cellar of her new restaurant. Sarah has a brilliant idea she will design an "electric skylight". In order to do that, she will install a solar panel on the roof, which can directly power three LED lights in the cellar with only wires connecting them. A long two core cable (positive and negative connection) connects the solar panel to the first LED light (distance c metres), and then two short lengths link two more LED lights (separated by distances d and e metres, all as marked on the diagram). The 1 sq mm cable used has a resistance of 0.03 2 per metre per core. The solar panel is modelled as a Norton equivalent source, while the LED lights are modelled as Thévenin voltage sources. Note: solar panels and LED lights are actually more complex than this, but assume that they behave this way unless you are given specific details about the components used by Sarah] c m e m c m Im e m Solar panel LED light 1 LED light 2 LED light 3 Figure 1. Sarah's design. As an enthusiastic future electrical engineer, Sarah asks for your help to analyse her proposed system when you visit. The values of the circuit elements and wire lengths have been emailed to you.

Answer 1

(c) The resistance per meter of wire is, For wire, c 28 m, therefore, c (28) (0.03)= 0.84 For wire, d 2.8 m, therefore, d (2.8)(0.03) 0.084 Q For wire, e 2.4 m, therefore, e (2.4) (0.03) = 0.072 Q r 0.03 /m - Bring down the top line to bottom to simplify the circuit 0.084 Ω ν. 0.84Q 0.072 LED3 LED2 SP 6.6 Q 6.6 Ω 6.6 Ω 0.92 A 560 Q 31.5 31.5 31.5 0.072 Q 0.084V 0.84 Q Convert the current source in solar panel to voltage source. V = IR = (0.92) (560)= 515.2 V Draw the modified circuit 0.84Q 0.072 2 0.084 LED3 LED2 SP 6.6 6.6 Ω 6.6 Q 560 Q 12 31.5 31.5 )31.5 515.2 V V 0.084 0.072 Q 0.84Q

Write the KVL equation for loop į. -515.2 5601 0.841 6.6( I,)+31.5+0.841 = 0 568.281-6.61, =483.7 (1) ...... Write the KVL equation for loop i, . -31.5+6.6(1, 1)+0.0841 0.0841, 6.6 (1,-1) + 31.5 = 0 -6.611 13.3681,-6.61,0 ...... (2) Write the KVL equation for loop i, . -31.5 6.6(1 -I,)+0.0721 6.61, +31.5 0.07213 = 0 -6.61,+13.3441, =0 (3) Solve the equations (1), (2) and (3) I 0.8577 A =0.5603 A I3 = 0.2771 A Calculate the power delivered by the solar panel. P (515.2)- () (560) -(515.2)(0.8577)-(0.8577)' (560) P 30 W Calculate the power dissipated by the LED 1 P= (31.5)(-1)(7,-1,)(6.6) (31.5)(0.8577-0.5603)+ (0.8577 -0.5603) (6.6) = = 9.368+0.5837 P1 = 9.952 W Calculate the power dissipated by the LED 2 Pi (315) (-)(,-1,)(6.6) (31.5)(0.5603-0.2771)+ (0.5603-0.2771) (6.6) = =8.92 0.5293 P, = 9.45 W

Calculate the power dissipated by the LED 3 P(31.5) ()) (6.6) =(31.5)(0.2771)+(0.2771) (6.6) = 8.729 0.5068 P3 = 9.236 W Calculate the total power dissipated in LED lights. PIEDPP2P3 -9.952 9.45+9.236 = 28.638 W P Calculate the percentage of transferred from the solar panel to the LED lights. P - x100% LED % power = 28.638 x 100% 30 % power = 95.46% Calculate the power dissipation in wiring. |(4 (0.84)+(4) (0.84)+(1, (0.084)+(1) (0.084) ,)' (0.072)+() (0.072) 2() (0.84)+2(,) (0.084)+2(1,) (0.072) =2 (0.8577) (0.84)+2(0.5603)' (0.084)+2(0.2771) (0.072) wiring -1.236 0.053 0.011 P 1.3 W wiring Total power dissipated in the circuit is, 28.6381.3 = 30 W

Hence, the total power dissipated across the element is equivalent to power at the source diss S Hence, the power is balanced in the system. (d) Draw the simulation diagram to calculate the currents 277.1mA 560.3mA 857.7mA R3 R1 R2 ww w ww 0.072 277.1mA 0.084 297.4mA 0.84 283.2mA R10 R9 R8 6.6 6.6 R7 6.6 560 297.4mA 283.2mA 277.1mA 857.7mA 31.5 31.5 V4 V1 515.2 31.5 857.7mA R4 R5 R6 857.7mA ww ww 0.84 0.084 0.072 560.3mA 277.1mA Draw the simulation diagram to cal cul ate the voltages R3 R2 R1 34.18V 33.40V 33.46V 33.42V 0.072 0.084 0.84 R10 R9 R8 6.6 6.6 6.6 R7 560 31.55V V2 V3 31.57V 31.5 514.5V 31.5 V4 515.2 31.5 R4 R5 R6 -720.4mV 67.01mV OV 47.06mV 0.84 0.084 0.072 ww

Draw the simulation diagram to calculate the voltages, currents and power 277.1mA 560.3mA 857.7mA R2 26.37MW R3 5.529mW R1617.9mW ww 34.18V ww 0.084 ww 33.40V 277.1mA 33.42V 33.46V 0.072 0.84 283.2mA 297.4mA R10 R9 R8 529.2mW 506.8mW 6.6 583.8mW 6.6 R7 411.9W 560 31.55V 297.4mA 9.369W 283.2mA 277.1mA 857.7mA 31.57V 8.919W 514.5V 31.5 31.5 V4 441.9W 515.2 31.5 8.729W 857.7mA R4 R5 26.37mW 5.529MW R6 857.7mA -720.4mV wW 47.06mV 67.01mV 0V 617.9mW 0.84 0.072 0.084 560.3mA 277.1mA Hence, the currents and power values are same as in part (c)