Respuesta :
Answer:
Bulb with power 25 W
Solution:
As per the question:
Voltage, V = 120 V
Power of bulb 1, P = 25 W
Power of bulb 2, P' = 100 W
Voltage in series, V' = 240 V
Now,
When the two bulbs are connected in parallel, voltage across them is same and the current in the two branches is different.
From the relation:
[tex]P = \frac{V^{2}}{R}[/tex]
Calculating the respective resistances of the two bulbs in each branch using the above relation:
[tex]25 = \frac{120^{2}}{R}[/tex]
[tex]R = \frac{120^{2}}{25} = 576 \Omega[/tex]
Similarly,
[tex]P' = \frac{V^{2}}{R'}[/tex]
[tex]100 = \frac{120^{2}}{R'}[/tex]
[tex]R' = \frac{120^{2}}{100} = 144 \Omega[/tex]
Now,
To calculate the maximum current in the respective branches:
[tex]I_{m} = \frac{V}{R}[/tex]
[tex]I_{m} = \frac{120}{576} = 0.208\ A[/tex]
[tex]I'_{m} = \frac{V}{R'} = \frac{120}{144} = 0.833[/tex]
Now,
When these two bulbs are connected in series across 240 V, then maximum current in the network:
[tex]I = \frac{V'}{R + R'} = \frac{240}{576 + 144} = 0.33 A[/tex]
Now, the bulb that will blow first in the series is the one with current rating less than 0.33 A.
Since in series the same current flows through each circuit element, thus the bulb 1 with power 25 W having current rating 0.208 A which is less than the current rating in series, i.e., 0.33 A will burn out.
