Pollination is the transfer of pollen for plant fertilization
The correct responses are as follows;
The description of the changes that take place is included in this response
6 (i) The circuit arrangement is a parallel circuit arrangement
(ii) The advantages are;
- Higher uptime of the bulbs
(iii) Lamp C glows the brightest
(iv) The total resistance is 4 Ω
The reason roe the above responses are as follows:
During the course of pollination, the pollen, through movement, finds it way to the germ-cell in the female ovary, where it fuses with the female egg (gamete) located inside the ovule
The resulting fertilization produces the zygote, which forms an embryo inside the ovule by dividing several times
A seed is formed from the ovule when it develops a tough coat, and by rapid growth a fruit is formed as the ovary matures
5 (i) The kind of combination, where all the wires that enters the bulb are joined on one side of the bulbs, and all the wires that comes from the bulbs are joined on the other end of the bulb, before joining the circuit is a parallel connection
(ii) The advantages of bulbs connected in parallel are;
- The voltage available in the power source reaches all the bulbs. That is each bulb had the voltage of the power source across it and therefore, is brighter than when the bulbs are arranged in series
- Other bulbs remain functional following the disconnection of one of the bulbs or loops
(iii) The bulb that glows brightest is given by the bulb that consumes the most energy per unit time, and therefore, the bulb that has the most power
Power consumed by each bulb is given as follows;
Bulb A, [tex]P_A[/tex] = 3 A × 60 V = 180 W
Bulb B, [tex]P_B[/tex] = 4 A × 60 V = 240 W
Bulb C, [tex]P_C[/tex] = 5 A × 60 V = 300 W
Bulb D, [tex]P_D[/tex] = 3 A × 60 V = 240 W
The highest power used by a bulb is 300 W, by bulb C
Therefore;
Bulb C is the brightest bulb
(iv) The total resistance of the circuit is given as follows;
[tex]Total \ resistance =\dfrac{Circuit \ voltage}{Total \ current}[/tex]
Total current, [tex]I_{tot}[/tex] = 3 A + 4 A + 5 A + 3 A = 15 A
Circuit voltage, V = 60 V (Given)
Therefore, we have;
[tex]Total \ resistance =\dfrac{60 \, V}{15 \, A} = 4 \, \Omega[/tex]
The total resistance is 4 Ω
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