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Answer :
Final Answer:
All bulbs glow with their full brightness because they each receive sufficient voltage and power. Thus, the correct answer is: a) All bulbs glow with their full brightness.
Explanation:
To determine the brightness of each bulb, we calculate the current passing through the circuit using Ohm's Law (V = IR). The total resistance (R_total) in the circuit is the sum of the resistances of each bulb connected in series.
[tex]R_total = R_1 + R_2 + R_3[/tex]
[tex]R_total[/tex] = [tex](V / P_1) + (V^2 / P_2) + (V^2 / P_3)[/tex]
[tex]R_total[/tex] = (220² / 25) + (220² / 60) + (220² / 100)
[tex]R_total[/tex] = 968 + 403.33 + 484
[tex]R_total[/tex] ≈ 1855.33 ohms
Then, we calculate the total current (I) passing through the circuit using Ohm's Law.
I = V / R_total
I = 220 / 1855.33
I ≈ 0.1187 A
Using the power formula (P = VI), we can find the power consumed by each bulb:
[tex]P_1 = VI_1[/tex]
[tex]P_1[/tex] = 220 * 0.1187
[tex]P_1[/tex] ≈ 26 W
[tex]P_2 = VI_2[/tex]
[tex]P_2[/tex] = 220 * 0.1187
[tex]P_2[/tex] ≈ 26 W
[tex]P_3 = VI_3[/tex]
[tex]P_3[/tex] = 220 * 0.1187
[tex]P_3[/tex] ≈ 26 W
Since the power ratings of all bulbs are less than the power supplied, they will all glow with their full brightness. Therefore, the correct answer is: a) All bulbs glow with their full brightness.
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